148979
An object, moving with velocity $5 \mathrm{~m} / \mathrm{s}$, undergoes an acceleration of $1 \mathrm{~m} / \mathrm{s}^{2}$ at time $t=$ 0. If the object has a mass of $1 \mathrm{~kg}$, the kinetic energy (KE) of the object at time $t=5 \mathrm{~s}$ is
1 $\mathrm{KE}=12.5$ joules
2 $\mathrm{KE}=20$ joules
3 $\mathrm{KE}=30$ joules
4 $\mathrm{KE}=50$ joules
5 $\mathrm{KE}=0$ joules
Explanation:
D Given, Initial velocity $(\mathrm{u})=5 \mathrm{~m} / \mathrm{s}$ Acceleration (a) $=1 \mathrm{~m} / \mathrm{s}^{2}$ Mass $(\mathrm{m})=1 \mathrm{~kg}$ Time $(\mathrm{t})=5 \mathrm{sec}$ From $1^{\text {st }}$ equation of motion, $\mathrm{v}=\mathrm{u}+\text { at }$ At $\mathrm{t}=5 \mathrm{sec}$, $\mathrm{v}=5+1 \times 5$ $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ Kinetic energy of the object (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \times 1 \times(10)^{2}$ $\mathrm{KE}=50$ joules
Kerala CEE-2019
Work, Energy and Power
148980
A particle of mass $5 \mathrm{~m}$ at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular direction with each speed $v$. The energy released during the process is
1 $\frac{3}{5} m v^{2}$
2 $\frac{5}{3} \mathrm{mv}^{2}$
3 $\frac{3}{2} \mathrm{mv}^{2}$
4 $\frac{4}{3} \mathrm{mv}^{2}$
Explanation:
D Linear momentum is conserved - $P_{i}=P_{f}$ $0=m v \hat{i}+m v \hat{j}+3 m \vec{v}^{\prime}$ $\vec{v}^{\prime}=\frac{-v \hat{i}-v \hat{j}}{3}$ $\left|\vec{v}^{\prime}\right|=\frac{1}{3} \sqrt{v^{2}+v^{2}}$ $\left|\vec{v}^{\prime}\right|=\frac{\sqrt{2}}{3} v$ Hence, $(\text { K.E. })_{\mathrm{f}} =\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(3 m) \frac{2}{9} v^{2}$ $=\frac{4 m v^{2}}{3}$ So, the energy released - $\text { K.E. }=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}} \quad \because \mathrm{KE}_{\mathrm{i}}=0$ $=\frac{4 \mathrm{mv}^{2}}{3}-0\left[\because(\mathrm{KE})_{\mathrm{i}}=0\right]$ $\text { K.E. }=\frac{4 \mathrm{mv}^{2}}{3}$
NEET (Odisha) - 2019
Work, Energy and Power
148982
The rate of decrease of kinetic energy is $9.6 \mathrm{~J} / \mathrm{s}$. Find the magnitude of force acting on particle when it's speed is $3 \mathrm{~m} / \mathrm{s}$.
1 $3.2 \mathrm{~N}$
2 $4.8 \mathrm{~N}$
3 $2.4 \mathrm{~N}$
4 $5.6 \mathrm{~N}$
Explanation:
A Rate of decrease of kinetic energy $\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right)=9.6 \mathrm{~J} / \mathrm{sec}$, speed $(\mathrm{v})=3 \mathrm{~m} / \mathrm{s}$ We know that, kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Rate of change of K.E. is $\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{1}{2} \mathrm{~m} \times 2 \mathrm{v} \times \frac{\mathrm{dv}}{\mathrm{dt}}$ $=\mathrm{v} \times \frac{\mathrm{mdv}}{\mathrm{dt}}$ $\frac{\mathrm{dK}}{\mathrm{dt}}=\mathrm{v} \times \text { Force }$ Force $=\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right) \times \frac{1}{\mathrm{v}}=\frac{9.6}{3}=3.2 \mathrm{~N}$
JIPMER-2019
Work, Energy and Power
148983
A body moves along a straight line and the variation of its kinetic energy with time is linear as shown in the figure below. Then the force acting on the body is
1 Zero
2 Constant greater than zero
3 Inversely proportional to velocity
4 Directly proportional to velocity
Explanation:
C As we know that, $\text { K.E. } \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2} \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{Ct}$ Differentiating w.r.t. time, we get, $\frac{\mathrm{m}}{2} \cdot 2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{C}$ $\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{v}}$ $\mathrm{F}=\frac{\mathrm{C}}{\mathrm{v}} \quad\left(\because \mathrm{F}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\right)$ $\mathrm{F} \propto \frac{1}{\mathrm{v}}$ So, force acting on body will be inversely proportional to velocity.
148979
An object, moving with velocity $5 \mathrm{~m} / \mathrm{s}$, undergoes an acceleration of $1 \mathrm{~m} / \mathrm{s}^{2}$ at time $t=$ 0. If the object has a mass of $1 \mathrm{~kg}$, the kinetic energy (KE) of the object at time $t=5 \mathrm{~s}$ is
1 $\mathrm{KE}=12.5$ joules
2 $\mathrm{KE}=20$ joules
3 $\mathrm{KE}=30$ joules
4 $\mathrm{KE}=50$ joules
5 $\mathrm{KE}=0$ joules
Explanation:
D Given, Initial velocity $(\mathrm{u})=5 \mathrm{~m} / \mathrm{s}$ Acceleration (a) $=1 \mathrm{~m} / \mathrm{s}^{2}$ Mass $(\mathrm{m})=1 \mathrm{~kg}$ Time $(\mathrm{t})=5 \mathrm{sec}$ From $1^{\text {st }}$ equation of motion, $\mathrm{v}=\mathrm{u}+\text { at }$ At $\mathrm{t}=5 \mathrm{sec}$, $\mathrm{v}=5+1 \times 5$ $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ Kinetic energy of the object (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \times 1 \times(10)^{2}$ $\mathrm{KE}=50$ joules
Kerala CEE-2019
Work, Energy and Power
148980
A particle of mass $5 \mathrm{~m}$ at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular direction with each speed $v$. The energy released during the process is
1 $\frac{3}{5} m v^{2}$
2 $\frac{5}{3} \mathrm{mv}^{2}$
3 $\frac{3}{2} \mathrm{mv}^{2}$
4 $\frac{4}{3} \mathrm{mv}^{2}$
Explanation:
D Linear momentum is conserved - $P_{i}=P_{f}$ $0=m v \hat{i}+m v \hat{j}+3 m \vec{v}^{\prime}$ $\vec{v}^{\prime}=\frac{-v \hat{i}-v \hat{j}}{3}$ $\left|\vec{v}^{\prime}\right|=\frac{1}{3} \sqrt{v^{2}+v^{2}}$ $\left|\vec{v}^{\prime}\right|=\frac{\sqrt{2}}{3} v$ Hence, $(\text { K.E. })_{\mathrm{f}} =\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(3 m) \frac{2}{9} v^{2}$ $=\frac{4 m v^{2}}{3}$ So, the energy released - $\text { K.E. }=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}} \quad \because \mathrm{KE}_{\mathrm{i}}=0$ $=\frac{4 \mathrm{mv}^{2}}{3}-0\left[\because(\mathrm{KE})_{\mathrm{i}}=0\right]$ $\text { K.E. }=\frac{4 \mathrm{mv}^{2}}{3}$
NEET (Odisha) - 2019
Work, Energy and Power
148982
The rate of decrease of kinetic energy is $9.6 \mathrm{~J} / \mathrm{s}$. Find the magnitude of force acting on particle when it's speed is $3 \mathrm{~m} / \mathrm{s}$.
1 $3.2 \mathrm{~N}$
2 $4.8 \mathrm{~N}$
3 $2.4 \mathrm{~N}$
4 $5.6 \mathrm{~N}$
Explanation:
A Rate of decrease of kinetic energy $\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right)=9.6 \mathrm{~J} / \mathrm{sec}$, speed $(\mathrm{v})=3 \mathrm{~m} / \mathrm{s}$ We know that, kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Rate of change of K.E. is $\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{1}{2} \mathrm{~m} \times 2 \mathrm{v} \times \frac{\mathrm{dv}}{\mathrm{dt}}$ $=\mathrm{v} \times \frac{\mathrm{mdv}}{\mathrm{dt}}$ $\frac{\mathrm{dK}}{\mathrm{dt}}=\mathrm{v} \times \text { Force }$ Force $=\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right) \times \frac{1}{\mathrm{v}}=\frac{9.6}{3}=3.2 \mathrm{~N}$
JIPMER-2019
Work, Energy and Power
148983
A body moves along a straight line and the variation of its kinetic energy with time is linear as shown in the figure below. Then the force acting on the body is
1 Zero
2 Constant greater than zero
3 Inversely proportional to velocity
4 Directly proportional to velocity
Explanation:
C As we know that, $\text { K.E. } \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2} \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{Ct}$ Differentiating w.r.t. time, we get, $\frac{\mathrm{m}}{2} \cdot 2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{C}$ $\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{v}}$ $\mathrm{F}=\frac{\mathrm{C}}{\mathrm{v}} \quad\left(\because \mathrm{F}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\right)$ $\mathrm{F} \propto \frac{1}{\mathrm{v}}$ So, force acting on body will be inversely proportional to velocity.
148979
An object, moving with velocity $5 \mathrm{~m} / \mathrm{s}$, undergoes an acceleration of $1 \mathrm{~m} / \mathrm{s}^{2}$ at time $t=$ 0. If the object has a mass of $1 \mathrm{~kg}$, the kinetic energy (KE) of the object at time $t=5 \mathrm{~s}$ is
1 $\mathrm{KE}=12.5$ joules
2 $\mathrm{KE}=20$ joules
3 $\mathrm{KE}=30$ joules
4 $\mathrm{KE}=50$ joules
5 $\mathrm{KE}=0$ joules
Explanation:
D Given, Initial velocity $(\mathrm{u})=5 \mathrm{~m} / \mathrm{s}$ Acceleration (a) $=1 \mathrm{~m} / \mathrm{s}^{2}$ Mass $(\mathrm{m})=1 \mathrm{~kg}$ Time $(\mathrm{t})=5 \mathrm{sec}$ From $1^{\text {st }}$ equation of motion, $\mathrm{v}=\mathrm{u}+\text { at }$ At $\mathrm{t}=5 \mathrm{sec}$, $\mathrm{v}=5+1 \times 5$ $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ Kinetic energy of the object (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \times 1 \times(10)^{2}$ $\mathrm{KE}=50$ joules
Kerala CEE-2019
Work, Energy and Power
148980
A particle of mass $5 \mathrm{~m}$ at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular direction with each speed $v$. The energy released during the process is
1 $\frac{3}{5} m v^{2}$
2 $\frac{5}{3} \mathrm{mv}^{2}$
3 $\frac{3}{2} \mathrm{mv}^{2}$
4 $\frac{4}{3} \mathrm{mv}^{2}$
Explanation:
D Linear momentum is conserved - $P_{i}=P_{f}$ $0=m v \hat{i}+m v \hat{j}+3 m \vec{v}^{\prime}$ $\vec{v}^{\prime}=\frac{-v \hat{i}-v \hat{j}}{3}$ $\left|\vec{v}^{\prime}\right|=\frac{1}{3} \sqrt{v^{2}+v^{2}}$ $\left|\vec{v}^{\prime}\right|=\frac{\sqrt{2}}{3} v$ Hence, $(\text { K.E. })_{\mathrm{f}} =\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(3 m) \frac{2}{9} v^{2}$ $=\frac{4 m v^{2}}{3}$ So, the energy released - $\text { K.E. }=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}} \quad \because \mathrm{KE}_{\mathrm{i}}=0$ $=\frac{4 \mathrm{mv}^{2}}{3}-0\left[\because(\mathrm{KE})_{\mathrm{i}}=0\right]$ $\text { K.E. }=\frac{4 \mathrm{mv}^{2}}{3}$
NEET (Odisha) - 2019
Work, Energy and Power
148982
The rate of decrease of kinetic energy is $9.6 \mathrm{~J} / \mathrm{s}$. Find the magnitude of force acting on particle when it's speed is $3 \mathrm{~m} / \mathrm{s}$.
1 $3.2 \mathrm{~N}$
2 $4.8 \mathrm{~N}$
3 $2.4 \mathrm{~N}$
4 $5.6 \mathrm{~N}$
Explanation:
A Rate of decrease of kinetic energy $\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right)=9.6 \mathrm{~J} / \mathrm{sec}$, speed $(\mathrm{v})=3 \mathrm{~m} / \mathrm{s}$ We know that, kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Rate of change of K.E. is $\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{1}{2} \mathrm{~m} \times 2 \mathrm{v} \times \frac{\mathrm{dv}}{\mathrm{dt}}$ $=\mathrm{v} \times \frac{\mathrm{mdv}}{\mathrm{dt}}$ $\frac{\mathrm{dK}}{\mathrm{dt}}=\mathrm{v} \times \text { Force }$ Force $=\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right) \times \frac{1}{\mathrm{v}}=\frac{9.6}{3}=3.2 \mathrm{~N}$
JIPMER-2019
Work, Energy and Power
148983
A body moves along a straight line and the variation of its kinetic energy with time is linear as shown in the figure below. Then the force acting on the body is
1 Zero
2 Constant greater than zero
3 Inversely proportional to velocity
4 Directly proportional to velocity
Explanation:
C As we know that, $\text { K.E. } \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2} \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{Ct}$ Differentiating w.r.t. time, we get, $\frac{\mathrm{m}}{2} \cdot 2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{C}$ $\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{v}}$ $\mathrm{F}=\frac{\mathrm{C}}{\mathrm{v}} \quad\left(\because \mathrm{F}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\right)$ $\mathrm{F} \propto \frac{1}{\mathrm{v}}$ So, force acting on body will be inversely proportional to velocity.
148979
An object, moving with velocity $5 \mathrm{~m} / \mathrm{s}$, undergoes an acceleration of $1 \mathrm{~m} / \mathrm{s}^{2}$ at time $t=$ 0. If the object has a mass of $1 \mathrm{~kg}$, the kinetic energy (KE) of the object at time $t=5 \mathrm{~s}$ is
1 $\mathrm{KE}=12.5$ joules
2 $\mathrm{KE}=20$ joules
3 $\mathrm{KE}=30$ joules
4 $\mathrm{KE}=50$ joules
5 $\mathrm{KE}=0$ joules
Explanation:
D Given, Initial velocity $(\mathrm{u})=5 \mathrm{~m} / \mathrm{s}$ Acceleration (a) $=1 \mathrm{~m} / \mathrm{s}^{2}$ Mass $(\mathrm{m})=1 \mathrm{~kg}$ Time $(\mathrm{t})=5 \mathrm{sec}$ From $1^{\text {st }}$ equation of motion, $\mathrm{v}=\mathrm{u}+\text { at }$ At $\mathrm{t}=5 \mathrm{sec}$, $\mathrm{v}=5+1 \times 5$ $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ Kinetic energy of the object (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \times 1 \times(10)^{2}$ $\mathrm{KE}=50$ joules
Kerala CEE-2019
Work, Energy and Power
148980
A particle of mass $5 \mathrm{~m}$ at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular direction with each speed $v$. The energy released during the process is
1 $\frac{3}{5} m v^{2}$
2 $\frac{5}{3} \mathrm{mv}^{2}$
3 $\frac{3}{2} \mathrm{mv}^{2}$
4 $\frac{4}{3} \mathrm{mv}^{2}$
Explanation:
D Linear momentum is conserved - $P_{i}=P_{f}$ $0=m v \hat{i}+m v \hat{j}+3 m \vec{v}^{\prime}$ $\vec{v}^{\prime}=\frac{-v \hat{i}-v \hat{j}}{3}$ $\left|\vec{v}^{\prime}\right|=\frac{1}{3} \sqrt{v^{2}+v^{2}}$ $\left|\vec{v}^{\prime}\right|=\frac{\sqrt{2}}{3} v$ Hence, $(\text { K.E. })_{\mathrm{f}} =\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(3 m) \frac{2}{9} v^{2}$ $=\frac{4 m v^{2}}{3}$ So, the energy released - $\text { K.E. }=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}} \quad \because \mathrm{KE}_{\mathrm{i}}=0$ $=\frac{4 \mathrm{mv}^{2}}{3}-0\left[\because(\mathrm{KE})_{\mathrm{i}}=0\right]$ $\text { K.E. }=\frac{4 \mathrm{mv}^{2}}{3}$
NEET (Odisha) - 2019
Work, Energy and Power
148982
The rate of decrease of kinetic energy is $9.6 \mathrm{~J} / \mathrm{s}$. Find the magnitude of force acting on particle when it's speed is $3 \mathrm{~m} / \mathrm{s}$.
1 $3.2 \mathrm{~N}$
2 $4.8 \mathrm{~N}$
3 $2.4 \mathrm{~N}$
4 $5.6 \mathrm{~N}$
Explanation:
A Rate of decrease of kinetic energy $\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right)=9.6 \mathrm{~J} / \mathrm{sec}$, speed $(\mathrm{v})=3 \mathrm{~m} / \mathrm{s}$ We know that, kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Rate of change of K.E. is $\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{1}{2} \mathrm{~m} \times 2 \mathrm{v} \times \frac{\mathrm{dv}}{\mathrm{dt}}$ $=\mathrm{v} \times \frac{\mathrm{mdv}}{\mathrm{dt}}$ $\frac{\mathrm{dK}}{\mathrm{dt}}=\mathrm{v} \times \text { Force }$ Force $=\left(\frac{\mathrm{dK}}{\mathrm{dt}}\right) \times \frac{1}{\mathrm{v}}=\frac{9.6}{3}=3.2 \mathrm{~N}$
JIPMER-2019
Work, Energy and Power
148983
A body moves along a straight line and the variation of its kinetic energy with time is linear as shown in the figure below. Then the force acting on the body is
1 Zero
2 Constant greater than zero
3 Inversely proportional to velocity
4 Directly proportional to velocity
Explanation:
C As we know that, $\text { K.E. } \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2} \propto \mathrm{t}$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{Ct}$ Differentiating w.r.t. time, we get, $\frac{\mathrm{m}}{2} \cdot 2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{C}$ $\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{v}}$ $\mathrm{F}=\frac{\mathrm{C}}{\mathrm{v}} \quad\left(\because \mathrm{F}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\right)$ $\mathrm{F} \propto \frac{1}{\mathrm{v}}$ So, force acting on body will be inversely proportional to velocity.
