149006
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
1 $x^{2}$
2 $\mathrm{e}^{\mathrm{x}}$
3 $\mathrm{x}$
4 $\log _{\mathrm{e}} \mathrm{x}$
Explanation:
A Given that, $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{a}=$ Acceleration $\mathrm{x}=$ displacement $\mathrm{k}=$ proportionality constant And negative sign is due to retardation. Now, from equation $\mathrm{a}=-\mathrm{kx}$ $\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{kx} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]$ $\frac{\mathrm{dv}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}$ $\mathrm{vdv}=-\mathrm{kx} \cdot \mathrm{dx}$ Taking integration both side, we get- $\int_{v_{1}}^{v_{2}} \mathrm{v} \cdot d v=\int-k x \cdot d x$ ${\left[\frac{v^{2}}{2}\right]_{v_{1}}^{v_{2}}=-k \cdot \frac{x^{2}}{2}}$ Multiplying by $\mathrm{m}$ both side, we get- $\frac{1}{2}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right) \mathrm{m}=-\frac{1}{2} \mathrm{kx}^{2} \times \mathrm{m}$ $\begin{aligned} \text { Loss in kinetic energy }(\Delta \mathrm{K} . \mathrm{E}) =-\frac{1}{2} \mathrm{mkx}^{2} \\ (\Delta \mathrm{K} . \mathrm{E}) \propto \mathrm{x}^{2}\end{aligned}$
UPSEE - 2007
Work, Energy and Power
149007
A steel ball of mass $5 \mathrm{~g}$ is thrown downward with velocity $10 \mathrm{~m} / \mathrm{s}$ from height $19.5 \mathrm{~m}$. It penetrates sand by $50 \mathrm{~cm}$. The change in mechanical energy will be: $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
149008
The potential energy for a force field $\vec{F}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is -
B Given that, Potential energy, $\mathrm{U}(\mathrm{x}, \mathrm{y})=\cos (\mathrm{x}+\mathrm{y})$ But we can represent a force field as, $\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}$ $\vec{F}=-\left[\frac{\partial U}{\partial x} \hat{i}+\frac{\partial U}{\partial y} \hat{j}\right]$ So, $\mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$ ${\left[\mathrm{F}_{\mathrm{x}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}}$ And $\quad F_{y}=-\frac{\partial U}{\partial y}=\sin (x+y)$ $\left[\mathrm{F}_{\mathrm{y}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}$ Putting the value of $F_{x}$ and $F_{y}$ in equation (i), we get- $\vec{F}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ $\vec{F}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
BITSAT-2010
Work, Energy and Power
149009
A body of mass $5.0 \mathrm{~kg}$ is moving with linear momentum $10 \mathrm{~kg}-\mathrm{ms}^{-1}$. A force of $0.2 \mathrm{~N}$ is applied on the body for $10 \mathrm{~s}$ in the direction of motion of body. The increase in kinetic energy of body will be
1 $2.8 \mathrm{~J}$
2 $3.2 \mathrm{~J}$
3 $3.6 \mathrm{~J}$
4 $4.4 \mathrm{~J}$
Explanation:
D Given that, Mass of body $(\mathrm{m})=5.0 \mathrm{~kg}$ Momentum (p) $=10 \mathrm{~kg}-\mathrm{ms}^{-1}$ Force $(F)=0.2 \mathrm{~N}$, Time $(\mathrm{t})=10 \mathrm{~s}$ $\because \quad \mathrm{p}=\mathrm{mu}$ $\therefore$ $\mathrm{u}=\frac{\mathrm{p}}{\mathrm{m}}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}$ From the first equation of motion- $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{v}=\mathrm{u}+\frac{\mathrm{F}}{\mathrm{m}} \mathrm{t} \quad\left(\because \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\right)$ $\mathrm{v}=2+\frac{0.2}{5} \times 10=2.4 \mathrm{~m} / \mathrm{s}$ The increase in kinetic energy $=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$ $=\frac{1}{2} \times 5\left[(2.4)^{2}-(2)^{2}\right]$ $=\frac{1}{2} \times 5[5.76-4]=4.4 \mathrm{~J}$
CG PET- 2011
Work, Energy and Power
149010
A frictionless track $A, B, C, D, E$ end in a circular loop of radius $R$. A body slides down the track from point $A$ which is at a height $h=$ $5 \mathrm{~cm}$. Maximum value of $R$ for the body to successful complete the loop is
1 $5 \mathrm{~cm}$
2 $\frac{15}{4} \mathrm{~cm}$
3 $\frac{10}{3} \mathrm{~cm}$
4 $2 \mathrm{~cm}$
Explanation:
D Given that, Height of track $(\mathrm{h})=5 \mathrm{~cm}$ Radius of circular loop $=\mathrm{R}$ Potential energy of the body at point $\mathrm{A}=$ Kinetic energy of the body at point $B$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{gh}=\frac{\mathrm{v}^{2}}{2}$ Velocity at lowest point in a vertical circle $\mathrm{v}=\sqrt{5 \mathrm{Rg}}$ Putting the value of $v$ in equation (i), we get- $\begin{aligned} \mathrm{gh} =\frac{5 \mathrm{Rg}}{2} \\ \mathrm{R} =\frac{2 \mathrm{~h}}{5} \quad[\because \mathrm{h}=5 \mathrm{~cm}] \\ \therefore \quad \mathrm{R} =\frac{2 \times 5}{5}=2 \mathrm{~cm}\end{aligned}$
149006
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
1 $x^{2}$
2 $\mathrm{e}^{\mathrm{x}}$
3 $\mathrm{x}$
4 $\log _{\mathrm{e}} \mathrm{x}$
Explanation:
A Given that, $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{a}=$ Acceleration $\mathrm{x}=$ displacement $\mathrm{k}=$ proportionality constant And negative sign is due to retardation. Now, from equation $\mathrm{a}=-\mathrm{kx}$ $\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{kx} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]$ $\frac{\mathrm{dv}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}$ $\mathrm{vdv}=-\mathrm{kx} \cdot \mathrm{dx}$ Taking integration both side, we get- $\int_{v_{1}}^{v_{2}} \mathrm{v} \cdot d v=\int-k x \cdot d x$ ${\left[\frac{v^{2}}{2}\right]_{v_{1}}^{v_{2}}=-k \cdot \frac{x^{2}}{2}}$ Multiplying by $\mathrm{m}$ both side, we get- $\frac{1}{2}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right) \mathrm{m}=-\frac{1}{2} \mathrm{kx}^{2} \times \mathrm{m}$ $\begin{aligned} \text { Loss in kinetic energy }(\Delta \mathrm{K} . \mathrm{E}) =-\frac{1}{2} \mathrm{mkx}^{2} \\ (\Delta \mathrm{K} . \mathrm{E}) \propto \mathrm{x}^{2}\end{aligned}$
UPSEE - 2007
Work, Energy and Power
149007
A steel ball of mass $5 \mathrm{~g}$ is thrown downward with velocity $10 \mathrm{~m} / \mathrm{s}$ from height $19.5 \mathrm{~m}$. It penetrates sand by $50 \mathrm{~cm}$. The change in mechanical energy will be: $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
149008
The potential energy for a force field $\vec{F}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is -
B Given that, Potential energy, $\mathrm{U}(\mathrm{x}, \mathrm{y})=\cos (\mathrm{x}+\mathrm{y})$ But we can represent a force field as, $\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}$ $\vec{F}=-\left[\frac{\partial U}{\partial x} \hat{i}+\frac{\partial U}{\partial y} \hat{j}\right]$ So, $\mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$ ${\left[\mathrm{F}_{\mathrm{x}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}}$ And $\quad F_{y}=-\frac{\partial U}{\partial y}=\sin (x+y)$ $\left[\mathrm{F}_{\mathrm{y}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}$ Putting the value of $F_{x}$ and $F_{y}$ in equation (i), we get- $\vec{F}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ $\vec{F}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
BITSAT-2010
Work, Energy and Power
149009
A body of mass $5.0 \mathrm{~kg}$ is moving with linear momentum $10 \mathrm{~kg}-\mathrm{ms}^{-1}$. A force of $0.2 \mathrm{~N}$ is applied on the body for $10 \mathrm{~s}$ in the direction of motion of body. The increase in kinetic energy of body will be
1 $2.8 \mathrm{~J}$
2 $3.2 \mathrm{~J}$
3 $3.6 \mathrm{~J}$
4 $4.4 \mathrm{~J}$
Explanation:
D Given that, Mass of body $(\mathrm{m})=5.0 \mathrm{~kg}$ Momentum (p) $=10 \mathrm{~kg}-\mathrm{ms}^{-1}$ Force $(F)=0.2 \mathrm{~N}$, Time $(\mathrm{t})=10 \mathrm{~s}$ $\because \quad \mathrm{p}=\mathrm{mu}$ $\therefore$ $\mathrm{u}=\frac{\mathrm{p}}{\mathrm{m}}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}$ From the first equation of motion- $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{v}=\mathrm{u}+\frac{\mathrm{F}}{\mathrm{m}} \mathrm{t} \quad\left(\because \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\right)$ $\mathrm{v}=2+\frac{0.2}{5} \times 10=2.4 \mathrm{~m} / \mathrm{s}$ The increase in kinetic energy $=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$ $=\frac{1}{2} \times 5\left[(2.4)^{2}-(2)^{2}\right]$ $=\frac{1}{2} \times 5[5.76-4]=4.4 \mathrm{~J}$
CG PET- 2011
Work, Energy and Power
149010
A frictionless track $A, B, C, D, E$ end in a circular loop of radius $R$. A body slides down the track from point $A$ which is at a height $h=$ $5 \mathrm{~cm}$. Maximum value of $R$ for the body to successful complete the loop is
1 $5 \mathrm{~cm}$
2 $\frac{15}{4} \mathrm{~cm}$
3 $\frac{10}{3} \mathrm{~cm}$
4 $2 \mathrm{~cm}$
Explanation:
D Given that, Height of track $(\mathrm{h})=5 \mathrm{~cm}$ Radius of circular loop $=\mathrm{R}$ Potential energy of the body at point $\mathrm{A}=$ Kinetic energy of the body at point $B$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{gh}=\frac{\mathrm{v}^{2}}{2}$ Velocity at lowest point in a vertical circle $\mathrm{v}=\sqrt{5 \mathrm{Rg}}$ Putting the value of $v$ in equation (i), we get- $\begin{aligned} \mathrm{gh} =\frac{5 \mathrm{Rg}}{2} \\ \mathrm{R} =\frac{2 \mathrm{~h}}{5} \quad[\because \mathrm{h}=5 \mathrm{~cm}] \\ \therefore \quad \mathrm{R} =\frac{2 \times 5}{5}=2 \mathrm{~cm}\end{aligned}$
149006
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
1 $x^{2}$
2 $\mathrm{e}^{\mathrm{x}}$
3 $\mathrm{x}$
4 $\log _{\mathrm{e}} \mathrm{x}$
Explanation:
A Given that, $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{a}=$ Acceleration $\mathrm{x}=$ displacement $\mathrm{k}=$ proportionality constant And negative sign is due to retardation. Now, from equation $\mathrm{a}=-\mathrm{kx}$ $\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{kx} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]$ $\frac{\mathrm{dv}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}$ $\mathrm{vdv}=-\mathrm{kx} \cdot \mathrm{dx}$ Taking integration both side, we get- $\int_{v_{1}}^{v_{2}} \mathrm{v} \cdot d v=\int-k x \cdot d x$ ${\left[\frac{v^{2}}{2}\right]_{v_{1}}^{v_{2}}=-k \cdot \frac{x^{2}}{2}}$ Multiplying by $\mathrm{m}$ both side, we get- $\frac{1}{2}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right) \mathrm{m}=-\frac{1}{2} \mathrm{kx}^{2} \times \mathrm{m}$ $\begin{aligned} \text { Loss in kinetic energy }(\Delta \mathrm{K} . \mathrm{E}) =-\frac{1}{2} \mathrm{mkx}^{2} \\ (\Delta \mathrm{K} . \mathrm{E}) \propto \mathrm{x}^{2}\end{aligned}$
UPSEE - 2007
Work, Energy and Power
149007
A steel ball of mass $5 \mathrm{~g}$ is thrown downward with velocity $10 \mathrm{~m} / \mathrm{s}$ from height $19.5 \mathrm{~m}$. It penetrates sand by $50 \mathrm{~cm}$. The change in mechanical energy will be: $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
149008
The potential energy for a force field $\vec{F}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is -
B Given that, Potential energy, $\mathrm{U}(\mathrm{x}, \mathrm{y})=\cos (\mathrm{x}+\mathrm{y})$ But we can represent a force field as, $\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}$ $\vec{F}=-\left[\frac{\partial U}{\partial x} \hat{i}+\frac{\partial U}{\partial y} \hat{j}\right]$ So, $\mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$ ${\left[\mathrm{F}_{\mathrm{x}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}}$ And $\quad F_{y}=-\frac{\partial U}{\partial y}=\sin (x+y)$ $\left[\mathrm{F}_{\mathrm{y}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}$ Putting the value of $F_{x}$ and $F_{y}$ in equation (i), we get- $\vec{F}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ $\vec{F}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
BITSAT-2010
Work, Energy and Power
149009
A body of mass $5.0 \mathrm{~kg}$ is moving with linear momentum $10 \mathrm{~kg}-\mathrm{ms}^{-1}$. A force of $0.2 \mathrm{~N}$ is applied on the body for $10 \mathrm{~s}$ in the direction of motion of body. The increase in kinetic energy of body will be
1 $2.8 \mathrm{~J}$
2 $3.2 \mathrm{~J}$
3 $3.6 \mathrm{~J}$
4 $4.4 \mathrm{~J}$
Explanation:
D Given that, Mass of body $(\mathrm{m})=5.0 \mathrm{~kg}$ Momentum (p) $=10 \mathrm{~kg}-\mathrm{ms}^{-1}$ Force $(F)=0.2 \mathrm{~N}$, Time $(\mathrm{t})=10 \mathrm{~s}$ $\because \quad \mathrm{p}=\mathrm{mu}$ $\therefore$ $\mathrm{u}=\frac{\mathrm{p}}{\mathrm{m}}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}$ From the first equation of motion- $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{v}=\mathrm{u}+\frac{\mathrm{F}}{\mathrm{m}} \mathrm{t} \quad\left(\because \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\right)$ $\mathrm{v}=2+\frac{0.2}{5} \times 10=2.4 \mathrm{~m} / \mathrm{s}$ The increase in kinetic energy $=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$ $=\frac{1}{2} \times 5\left[(2.4)^{2}-(2)^{2}\right]$ $=\frac{1}{2} \times 5[5.76-4]=4.4 \mathrm{~J}$
CG PET- 2011
Work, Energy and Power
149010
A frictionless track $A, B, C, D, E$ end in a circular loop of radius $R$. A body slides down the track from point $A$ which is at a height $h=$ $5 \mathrm{~cm}$. Maximum value of $R$ for the body to successful complete the loop is
1 $5 \mathrm{~cm}$
2 $\frac{15}{4} \mathrm{~cm}$
3 $\frac{10}{3} \mathrm{~cm}$
4 $2 \mathrm{~cm}$
Explanation:
D Given that, Height of track $(\mathrm{h})=5 \mathrm{~cm}$ Radius of circular loop $=\mathrm{R}$ Potential energy of the body at point $\mathrm{A}=$ Kinetic energy of the body at point $B$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{gh}=\frac{\mathrm{v}^{2}}{2}$ Velocity at lowest point in a vertical circle $\mathrm{v}=\sqrt{5 \mathrm{Rg}}$ Putting the value of $v$ in equation (i), we get- $\begin{aligned} \mathrm{gh} =\frac{5 \mathrm{Rg}}{2} \\ \mathrm{R} =\frac{2 \mathrm{~h}}{5} \quad[\because \mathrm{h}=5 \mathrm{~cm}] \\ \therefore \quad \mathrm{R} =\frac{2 \times 5}{5}=2 \mathrm{~cm}\end{aligned}$
149006
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
1 $x^{2}$
2 $\mathrm{e}^{\mathrm{x}}$
3 $\mathrm{x}$
4 $\log _{\mathrm{e}} \mathrm{x}$
Explanation:
A Given that, $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{a}=$ Acceleration $\mathrm{x}=$ displacement $\mathrm{k}=$ proportionality constant And negative sign is due to retardation. Now, from equation $\mathrm{a}=-\mathrm{kx}$ $\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{kx} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]$ $\frac{\mathrm{dv}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}$ $\mathrm{vdv}=-\mathrm{kx} \cdot \mathrm{dx}$ Taking integration both side, we get- $\int_{v_{1}}^{v_{2}} \mathrm{v} \cdot d v=\int-k x \cdot d x$ ${\left[\frac{v^{2}}{2}\right]_{v_{1}}^{v_{2}}=-k \cdot \frac{x^{2}}{2}}$ Multiplying by $\mathrm{m}$ both side, we get- $\frac{1}{2}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right) \mathrm{m}=-\frac{1}{2} \mathrm{kx}^{2} \times \mathrm{m}$ $\begin{aligned} \text { Loss in kinetic energy }(\Delta \mathrm{K} . \mathrm{E}) =-\frac{1}{2} \mathrm{mkx}^{2} \\ (\Delta \mathrm{K} . \mathrm{E}) \propto \mathrm{x}^{2}\end{aligned}$
UPSEE - 2007
Work, Energy and Power
149007
A steel ball of mass $5 \mathrm{~g}$ is thrown downward with velocity $10 \mathrm{~m} / \mathrm{s}$ from height $19.5 \mathrm{~m}$. It penetrates sand by $50 \mathrm{~cm}$. The change in mechanical energy will be: $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
149008
The potential energy for a force field $\vec{F}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is -
B Given that, Potential energy, $\mathrm{U}(\mathrm{x}, \mathrm{y})=\cos (\mathrm{x}+\mathrm{y})$ But we can represent a force field as, $\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}$ $\vec{F}=-\left[\frac{\partial U}{\partial x} \hat{i}+\frac{\partial U}{\partial y} \hat{j}\right]$ So, $\mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$ ${\left[\mathrm{F}_{\mathrm{x}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}}$ And $\quad F_{y}=-\frac{\partial U}{\partial y}=\sin (x+y)$ $\left[\mathrm{F}_{\mathrm{y}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}$ Putting the value of $F_{x}$ and $F_{y}$ in equation (i), we get- $\vec{F}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ $\vec{F}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
BITSAT-2010
Work, Energy and Power
149009
A body of mass $5.0 \mathrm{~kg}$ is moving with linear momentum $10 \mathrm{~kg}-\mathrm{ms}^{-1}$. A force of $0.2 \mathrm{~N}$ is applied on the body for $10 \mathrm{~s}$ in the direction of motion of body. The increase in kinetic energy of body will be
1 $2.8 \mathrm{~J}$
2 $3.2 \mathrm{~J}$
3 $3.6 \mathrm{~J}$
4 $4.4 \mathrm{~J}$
Explanation:
D Given that, Mass of body $(\mathrm{m})=5.0 \mathrm{~kg}$ Momentum (p) $=10 \mathrm{~kg}-\mathrm{ms}^{-1}$ Force $(F)=0.2 \mathrm{~N}$, Time $(\mathrm{t})=10 \mathrm{~s}$ $\because \quad \mathrm{p}=\mathrm{mu}$ $\therefore$ $\mathrm{u}=\frac{\mathrm{p}}{\mathrm{m}}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}$ From the first equation of motion- $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{v}=\mathrm{u}+\frac{\mathrm{F}}{\mathrm{m}} \mathrm{t} \quad\left(\because \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\right)$ $\mathrm{v}=2+\frac{0.2}{5} \times 10=2.4 \mathrm{~m} / \mathrm{s}$ The increase in kinetic energy $=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$ $=\frac{1}{2} \times 5\left[(2.4)^{2}-(2)^{2}\right]$ $=\frac{1}{2} \times 5[5.76-4]=4.4 \mathrm{~J}$
CG PET- 2011
Work, Energy and Power
149010
A frictionless track $A, B, C, D, E$ end in a circular loop of radius $R$. A body slides down the track from point $A$ which is at a height $h=$ $5 \mathrm{~cm}$. Maximum value of $R$ for the body to successful complete the loop is
1 $5 \mathrm{~cm}$
2 $\frac{15}{4} \mathrm{~cm}$
3 $\frac{10}{3} \mathrm{~cm}$
4 $2 \mathrm{~cm}$
Explanation:
D Given that, Height of track $(\mathrm{h})=5 \mathrm{~cm}$ Radius of circular loop $=\mathrm{R}$ Potential energy of the body at point $\mathrm{A}=$ Kinetic energy of the body at point $B$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{gh}=\frac{\mathrm{v}^{2}}{2}$ Velocity at lowest point in a vertical circle $\mathrm{v}=\sqrt{5 \mathrm{Rg}}$ Putting the value of $v$ in equation (i), we get- $\begin{aligned} \mathrm{gh} =\frac{5 \mathrm{Rg}}{2} \\ \mathrm{R} =\frac{2 \mathrm{~h}}{5} \quad[\because \mathrm{h}=5 \mathrm{~cm}] \\ \therefore \quad \mathrm{R} =\frac{2 \times 5}{5}=2 \mathrm{~cm}\end{aligned}$
149006
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
1 $x^{2}$
2 $\mathrm{e}^{\mathrm{x}}$
3 $\mathrm{x}$
4 $\log _{\mathrm{e}} \mathrm{x}$
Explanation:
A Given that, $\mathrm{a}=-\mathrm{kx}$ Where, $\mathrm{a}=$ Acceleration $\mathrm{x}=$ displacement $\mathrm{k}=$ proportionality constant And negative sign is due to retardation. Now, from equation $\mathrm{a}=-\mathrm{kx}$ $\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{kx} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]$ $\frac{\mathrm{dv}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}$ $\mathrm{vdv}=-\mathrm{kx} \cdot \mathrm{dx}$ Taking integration both side, we get- $\int_{v_{1}}^{v_{2}} \mathrm{v} \cdot d v=\int-k x \cdot d x$ ${\left[\frac{v^{2}}{2}\right]_{v_{1}}^{v_{2}}=-k \cdot \frac{x^{2}}{2}}$ Multiplying by $\mathrm{m}$ both side, we get- $\frac{1}{2}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right) \mathrm{m}=-\frac{1}{2} \mathrm{kx}^{2} \times \mathrm{m}$ $\begin{aligned} \text { Loss in kinetic energy }(\Delta \mathrm{K} . \mathrm{E}) =-\frac{1}{2} \mathrm{mkx}^{2} \\ (\Delta \mathrm{K} . \mathrm{E}) \propto \mathrm{x}^{2}\end{aligned}$
UPSEE - 2007
Work, Energy and Power
149007
A steel ball of mass $5 \mathrm{~g}$ is thrown downward with velocity $10 \mathrm{~m} / \mathrm{s}$ from height $19.5 \mathrm{~m}$. It penetrates sand by $50 \mathrm{~cm}$. The change in mechanical energy will be: $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
149008
The potential energy for a force field $\vec{F}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is -
B Given that, Potential energy, $\mathrm{U}(\mathrm{x}, \mathrm{y})=\cos (\mathrm{x}+\mathrm{y})$ But we can represent a force field as, $\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}$ $\vec{F}=-\left[\frac{\partial U}{\partial x} \hat{i}+\frac{\partial U}{\partial y} \hat{j}\right]$ So, $\mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$ ${\left[\mathrm{F}_{\mathrm{x}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}}$ And $\quad F_{y}=-\frac{\partial U}{\partial y}=\sin (x+y)$ $\left[\mathrm{F}_{\mathrm{y}}\right]_{\left(0, \frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}$ Putting the value of $F_{x}$ and $F_{y}$ in equation (i), we get- $\vec{F}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ $\vec{F}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
BITSAT-2010
Work, Energy and Power
149009
A body of mass $5.0 \mathrm{~kg}$ is moving with linear momentum $10 \mathrm{~kg}-\mathrm{ms}^{-1}$. A force of $0.2 \mathrm{~N}$ is applied on the body for $10 \mathrm{~s}$ in the direction of motion of body. The increase in kinetic energy of body will be
1 $2.8 \mathrm{~J}$
2 $3.2 \mathrm{~J}$
3 $3.6 \mathrm{~J}$
4 $4.4 \mathrm{~J}$
Explanation:
D Given that, Mass of body $(\mathrm{m})=5.0 \mathrm{~kg}$ Momentum (p) $=10 \mathrm{~kg}-\mathrm{ms}^{-1}$ Force $(F)=0.2 \mathrm{~N}$, Time $(\mathrm{t})=10 \mathrm{~s}$ $\because \quad \mathrm{p}=\mathrm{mu}$ $\therefore$ $\mathrm{u}=\frac{\mathrm{p}}{\mathrm{m}}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}$ From the first equation of motion- $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{v}=\mathrm{u}+\frac{\mathrm{F}}{\mathrm{m}} \mathrm{t} \quad\left(\because \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\right)$ $\mathrm{v}=2+\frac{0.2}{5} \times 10=2.4 \mathrm{~m} / \mathrm{s}$ The increase in kinetic energy $=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$ $=\frac{1}{2} \times 5\left[(2.4)^{2}-(2)^{2}\right]$ $=\frac{1}{2} \times 5[5.76-4]=4.4 \mathrm{~J}$
CG PET- 2011
Work, Energy and Power
149010
A frictionless track $A, B, C, D, E$ end in a circular loop of radius $R$. A body slides down the track from point $A$ which is at a height $h=$ $5 \mathrm{~cm}$. Maximum value of $R$ for the body to successful complete the loop is
1 $5 \mathrm{~cm}$
2 $\frac{15}{4} \mathrm{~cm}$
3 $\frac{10}{3} \mathrm{~cm}$
4 $2 \mathrm{~cm}$
Explanation:
D Given that, Height of track $(\mathrm{h})=5 \mathrm{~cm}$ Radius of circular loop $=\mathrm{R}$ Potential energy of the body at point $\mathrm{A}=$ Kinetic energy of the body at point $B$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{gh}=\frac{\mathrm{v}^{2}}{2}$ Velocity at lowest point in a vertical circle $\mathrm{v}=\sqrt{5 \mathrm{Rg}}$ Putting the value of $v$ in equation (i), we get- $\begin{aligned} \mathrm{gh} =\frac{5 \mathrm{Rg}}{2} \\ \mathrm{R} =\frac{2 \mathrm{~h}}{5} \quad[\because \mathrm{h}=5 \mathrm{~cm}] \\ \therefore \quad \mathrm{R} =\frac{2 \times 5}{5}=2 \mathrm{~cm}\end{aligned}$
