355821
A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to ( \({S=}\) distance covered)
1 \({\ln (S)}\)
2 \({\log (S)}\)
3 \({\sqrt{S}}\)
4 \({\log \dfrac{1}{S}}\)
Explanation:
Force \({\propto \dfrac{1}{\text { Distance }}}\) \(F=\dfrac{k}{S} \quad[k=\text { constant }]\) Work done in moving a particle by small distance \({d s}\) is \(d W=F d S \Rightarrow d W=\dfrac{k}{S} d S\) Total work done will be, \({\int d W=\int \dfrac{k}{S} d S}\)\(W=k \ln (S)\). So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355822
When a spring is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x^{2}+b x^{3}\) where \(a\) and \(b\) are constants. The work done in stretching the unstretched rubberb and by \(L\) is
Work done in stretching the rubber band \(\begin{aligned}W & =\int_{0}^{L}\left(a x^{2}+b x^{3}\right) d x \\& =\dfrac{a L^{3}}{3}+\dfrac{b L^{4}}{4}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355823
A 10 \(kg\) mass moves along \(x\)-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from \(x = 0\) to \(x = 8\,cm\) ?
1 \(8 \times 10^{-2}\) Joules
2 \(16 \times 10^{-2}\) Joules
3 \(4 \times 10^{-4}\) Joules
4 \(1.6 \times 10^{-3}\) Joules
Explanation:
Work done \(=\) Mass \(\times\) Area covered in between acceleration displacement curve and displacement axis. \(=10 \times \dfrac{1}{2}\left(8 \times 10^{-2} \times 20 \times 10^{-2}\right)=8 \times 10^{-2} J .\)
PHXI06:WORK ENERGY AND POWER
355824
A force \(\vec{F}=(3 x y-5 z) \hat{j}+4 z \hat{k}\) is applied on a particle. The work done by the force when the particle moves from point \((0,0,0)\) to point \((2,4,0)\) as shown in figure.
1 \(\dfrac{280}{5}\) units
2 \(\dfrac{140}{5}\) units
3 \(\dfrac{232}{5}\) units
4 \(\dfrac{192}{5}\) units
Explanation:
The \(z\)-component of the force and the \(x\) component of displacement are ineffective here. \(d W=F_{x} d x+F_{y} d y+F_{z} d z=F_{y} d y=3 x y . d y\) Where \({F_z} = 0,dz = 0\& {F_x} = 0\) \(\begin{gathered}d W=6 x^{4} d x \Rightarrow W=\int_{0}^{2} 6 x^{4} d x \\W=192 / 5 .\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355825
A block of mass 1 \(kg\) is pulled along the curved path \(ACB\) by a tangential force as shown in figure. The work done by the frictional force when the block moves from \(A\) to \(B\) is
1 5 \(J\)
2 10 \(J\)
3 20 \(J\)
4 45 \(J\)
Explanation:
Consider a small segment on the curve. Let \(\Delta S\) is the length of the segment Small work done by the external force is \(\begin{aligned}& \Delta W=F \Delta S \\& F=m g \sin \theta+\mu m g \cos \theta \\& \Delta W=m g \Delta S \sin \theta+\mu m g \Delta S \cos \theta \\& \Delta W=m g \Delta y+\mu m g \Delta x \\& W=\int_{y_{A}}^{y_{B}} m g \Delta y+\mu m g \int_{x_{A}}^{x_{A}} \Delta x \\& W=m g(0)+\mu m g\left(x_{B}-x_{A}\right) \\& W=\mu m g(10 m) \\& W=0.2 \times 1 \times 10(10)=20 \mathrm{~J}\end{aligned}\)
355821
A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to ( \({S=}\) distance covered)
1 \({\ln (S)}\)
2 \({\log (S)}\)
3 \({\sqrt{S}}\)
4 \({\log \dfrac{1}{S}}\)
Explanation:
Force \({\propto \dfrac{1}{\text { Distance }}}\) \(F=\dfrac{k}{S} \quad[k=\text { constant }]\) Work done in moving a particle by small distance \({d s}\) is \(d W=F d S \Rightarrow d W=\dfrac{k}{S} d S\) Total work done will be, \({\int d W=\int \dfrac{k}{S} d S}\)\(W=k \ln (S)\). So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355822
When a spring is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x^{2}+b x^{3}\) where \(a\) and \(b\) are constants. The work done in stretching the unstretched rubberb and by \(L\) is
Work done in stretching the rubber band \(\begin{aligned}W & =\int_{0}^{L}\left(a x^{2}+b x^{3}\right) d x \\& =\dfrac{a L^{3}}{3}+\dfrac{b L^{4}}{4}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355823
A 10 \(kg\) mass moves along \(x\)-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from \(x = 0\) to \(x = 8\,cm\) ?
1 \(8 \times 10^{-2}\) Joules
2 \(16 \times 10^{-2}\) Joules
3 \(4 \times 10^{-4}\) Joules
4 \(1.6 \times 10^{-3}\) Joules
Explanation:
Work done \(=\) Mass \(\times\) Area covered in between acceleration displacement curve and displacement axis. \(=10 \times \dfrac{1}{2}\left(8 \times 10^{-2} \times 20 \times 10^{-2}\right)=8 \times 10^{-2} J .\)
PHXI06:WORK ENERGY AND POWER
355824
A force \(\vec{F}=(3 x y-5 z) \hat{j}+4 z \hat{k}\) is applied on a particle. The work done by the force when the particle moves from point \((0,0,0)\) to point \((2,4,0)\) as shown in figure.
1 \(\dfrac{280}{5}\) units
2 \(\dfrac{140}{5}\) units
3 \(\dfrac{232}{5}\) units
4 \(\dfrac{192}{5}\) units
Explanation:
The \(z\)-component of the force and the \(x\) component of displacement are ineffective here. \(d W=F_{x} d x+F_{y} d y+F_{z} d z=F_{y} d y=3 x y . d y\) Where \({F_z} = 0,dz = 0\& {F_x} = 0\) \(\begin{gathered}d W=6 x^{4} d x \Rightarrow W=\int_{0}^{2} 6 x^{4} d x \\W=192 / 5 .\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355825
A block of mass 1 \(kg\) is pulled along the curved path \(ACB\) by a tangential force as shown in figure. The work done by the frictional force when the block moves from \(A\) to \(B\) is
1 5 \(J\)
2 10 \(J\)
3 20 \(J\)
4 45 \(J\)
Explanation:
Consider a small segment on the curve. Let \(\Delta S\) is the length of the segment Small work done by the external force is \(\begin{aligned}& \Delta W=F \Delta S \\& F=m g \sin \theta+\mu m g \cos \theta \\& \Delta W=m g \Delta S \sin \theta+\mu m g \Delta S \cos \theta \\& \Delta W=m g \Delta y+\mu m g \Delta x \\& W=\int_{y_{A}}^{y_{B}} m g \Delta y+\mu m g \int_{x_{A}}^{x_{A}} \Delta x \\& W=m g(0)+\mu m g\left(x_{B}-x_{A}\right) \\& W=\mu m g(10 m) \\& W=0.2 \times 1 \times 10(10)=20 \mathrm{~J}\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355821
A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to ( \({S=}\) distance covered)
1 \({\ln (S)}\)
2 \({\log (S)}\)
3 \({\sqrt{S}}\)
4 \({\log \dfrac{1}{S}}\)
Explanation:
Force \({\propto \dfrac{1}{\text { Distance }}}\) \(F=\dfrac{k}{S} \quad[k=\text { constant }]\) Work done in moving a particle by small distance \({d s}\) is \(d W=F d S \Rightarrow d W=\dfrac{k}{S} d S\) Total work done will be, \({\int d W=\int \dfrac{k}{S} d S}\)\(W=k \ln (S)\). So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355822
When a spring is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x^{2}+b x^{3}\) where \(a\) and \(b\) are constants. The work done in stretching the unstretched rubberb and by \(L\) is
Work done in stretching the rubber band \(\begin{aligned}W & =\int_{0}^{L}\left(a x^{2}+b x^{3}\right) d x \\& =\dfrac{a L^{3}}{3}+\dfrac{b L^{4}}{4}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355823
A 10 \(kg\) mass moves along \(x\)-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from \(x = 0\) to \(x = 8\,cm\) ?
1 \(8 \times 10^{-2}\) Joules
2 \(16 \times 10^{-2}\) Joules
3 \(4 \times 10^{-4}\) Joules
4 \(1.6 \times 10^{-3}\) Joules
Explanation:
Work done \(=\) Mass \(\times\) Area covered in between acceleration displacement curve and displacement axis. \(=10 \times \dfrac{1}{2}\left(8 \times 10^{-2} \times 20 \times 10^{-2}\right)=8 \times 10^{-2} J .\)
PHXI06:WORK ENERGY AND POWER
355824
A force \(\vec{F}=(3 x y-5 z) \hat{j}+4 z \hat{k}\) is applied on a particle. The work done by the force when the particle moves from point \((0,0,0)\) to point \((2,4,0)\) as shown in figure.
1 \(\dfrac{280}{5}\) units
2 \(\dfrac{140}{5}\) units
3 \(\dfrac{232}{5}\) units
4 \(\dfrac{192}{5}\) units
Explanation:
The \(z\)-component of the force and the \(x\) component of displacement are ineffective here. \(d W=F_{x} d x+F_{y} d y+F_{z} d z=F_{y} d y=3 x y . d y\) Where \({F_z} = 0,dz = 0\& {F_x} = 0\) \(\begin{gathered}d W=6 x^{4} d x \Rightarrow W=\int_{0}^{2} 6 x^{4} d x \\W=192 / 5 .\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355825
A block of mass 1 \(kg\) is pulled along the curved path \(ACB\) by a tangential force as shown in figure. The work done by the frictional force when the block moves from \(A\) to \(B\) is
1 5 \(J\)
2 10 \(J\)
3 20 \(J\)
4 45 \(J\)
Explanation:
Consider a small segment on the curve. Let \(\Delta S\) is the length of the segment Small work done by the external force is \(\begin{aligned}& \Delta W=F \Delta S \\& F=m g \sin \theta+\mu m g \cos \theta \\& \Delta W=m g \Delta S \sin \theta+\mu m g \Delta S \cos \theta \\& \Delta W=m g \Delta y+\mu m g \Delta x \\& W=\int_{y_{A}}^{y_{B}} m g \Delta y+\mu m g \int_{x_{A}}^{x_{A}} \Delta x \\& W=m g(0)+\mu m g\left(x_{B}-x_{A}\right) \\& W=\mu m g(10 m) \\& W=0.2 \times 1 \times 10(10)=20 \mathrm{~J}\end{aligned}\)
355821
A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to ( \({S=}\) distance covered)
1 \({\ln (S)}\)
2 \({\log (S)}\)
3 \({\sqrt{S}}\)
4 \({\log \dfrac{1}{S}}\)
Explanation:
Force \({\propto \dfrac{1}{\text { Distance }}}\) \(F=\dfrac{k}{S} \quad[k=\text { constant }]\) Work done in moving a particle by small distance \({d s}\) is \(d W=F d S \Rightarrow d W=\dfrac{k}{S} d S\) Total work done will be, \({\int d W=\int \dfrac{k}{S} d S}\)\(W=k \ln (S)\). So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355822
When a spring is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x^{2}+b x^{3}\) where \(a\) and \(b\) are constants. The work done in stretching the unstretched rubberb and by \(L\) is
Work done in stretching the rubber band \(\begin{aligned}W & =\int_{0}^{L}\left(a x^{2}+b x^{3}\right) d x \\& =\dfrac{a L^{3}}{3}+\dfrac{b L^{4}}{4}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355823
A 10 \(kg\) mass moves along \(x\)-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from \(x = 0\) to \(x = 8\,cm\) ?
1 \(8 \times 10^{-2}\) Joules
2 \(16 \times 10^{-2}\) Joules
3 \(4 \times 10^{-4}\) Joules
4 \(1.6 \times 10^{-3}\) Joules
Explanation:
Work done \(=\) Mass \(\times\) Area covered in between acceleration displacement curve and displacement axis. \(=10 \times \dfrac{1}{2}\left(8 \times 10^{-2} \times 20 \times 10^{-2}\right)=8 \times 10^{-2} J .\)
PHXI06:WORK ENERGY AND POWER
355824
A force \(\vec{F}=(3 x y-5 z) \hat{j}+4 z \hat{k}\) is applied on a particle. The work done by the force when the particle moves from point \((0,0,0)\) to point \((2,4,0)\) as shown in figure.
1 \(\dfrac{280}{5}\) units
2 \(\dfrac{140}{5}\) units
3 \(\dfrac{232}{5}\) units
4 \(\dfrac{192}{5}\) units
Explanation:
The \(z\)-component of the force and the \(x\) component of displacement are ineffective here. \(d W=F_{x} d x+F_{y} d y+F_{z} d z=F_{y} d y=3 x y . d y\) Where \({F_z} = 0,dz = 0\& {F_x} = 0\) \(\begin{gathered}d W=6 x^{4} d x \Rightarrow W=\int_{0}^{2} 6 x^{4} d x \\W=192 / 5 .\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355825
A block of mass 1 \(kg\) is pulled along the curved path \(ACB\) by a tangential force as shown in figure. The work done by the frictional force when the block moves from \(A\) to \(B\) is
1 5 \(J\)
2 10 \(J\)
3 20 \(J\)
4 45 \(J\)
Explanation:
Consider a small segment on the curve. Let \(\Delta S\) is the length of the segment Small work done by the external force is \(\begin{aligned}& \Delta W=F \Delta S \\& F=m g \sin \theta+\mu m g \cos \theta \\& \Delta W=m g \Delta S \sin \theta+\mu m g \Delta S \cos \theta \\& \Delta W=m g \Delta y+\mu m g \Delta x \\& W=\int_{y_{A}}^{y_{B}} m g \Delta y+\mu m g \int_{x_{A}}^{x_{A}} \Delta x \\& W=m g(0)+\mu m g\left(x_{B}-x_{A}\right) \\& W=\mu m g(10 m) \\& W=0.2 \times 1 \times 10(10)=20 \mathrm{~J}\end{aligned}\)
355821
A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to ( \({S=}\) distance covered)
1 \({\ln (S)}\)
2 \({\log (S)}\)
3 \({\sqrt{S}}\)
4 \({\log \dfrac{1}{S}}\)
Explanation:
Force \({\propto \dfrac{1}{\text { Distance }}}\) \(F=\dfrac{k}{S} \quad[k=\text { constant }]\) Work done in moving a particle by small distance \({d s}\) is \(d W=F d S \Rightarrow d W=\dfrac{k}{S} d S\) Total work done will be, \({\int d W=\int \dfrac{k}{S} d S}\)\(W=k \ln (S)\). So correct option is (1)
PHXI06:WORK ENERGY AND POWER
355822
When a spring is stretched by a distance \(x\), it exerts a restoring force of magnitude \(F=a x^{2}+b x^{3}\) where \(a\) and \(b\) are constants. The work done in stretching the unstretched rubberb and by \(L\) is
Work done in stretching the rubber band \(\begin{aligned}W & =\int_{0}^{L}\left(a x^{2}+b x^{3}\right) d x \\& =\dfrac{a L^{3}}{3}+\dfrac{b L^{4}}{4}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355823
A 10 \(kg\) mass moves along \(x\)-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from \(x = 0\) to \(x = 8\,cm\) ?
1 \(8 \times 10^{-2}\) Joules
2 \(16 \times 10^{-2}\) Joules
3 \(4 \times 10^{-4}\) Joules
4 \(1.6 \times 10^{-3}\) Joules
Explanation:
Work done \(=\) Mass \(\times\) Area covered in between acceleration displacement curve and displacement axis. \(=10 \times \dfrac{1}{2}\left(8 \times 10^{-2} \times 20 \times 10^{-2}\right)=8 \times 10^{-2} J .\)
PHXI06:WORK ENERGY AND POWER
355824
A force \(\vec{F}=(3 x y-5 z) \hat{j}+4 z \hat{k}\) is applied on a particle. The work done by the force when the particle moves from point \((0,0,0)\) to point \((2,4,0)\) as shown in figure.
1 \(\dfrac{280}{5}\) units
2 \(\dfrac{140}{5}\) units
3 \(\dfrac{232}{5}\) units
4 \(\dfrac{192}{5}\) units
Explanation:
The \(z\)-component of the force and the \(x\) component of displacement are ineffective here. \(d W=F_{x} d x+F_{y} d y+F_{z} d z=F_{y} d y=3 x y . d y\) Where \({F_z} = 0,dz = 0\& {F_x} = 0\) \(\begin{gathered}d W=6 x^{4} d x \Rightarrow W=\int_{0}^{2} 6 x^{4} d x \\W=192 / 5 .\end{gathered}\)
PHXI06:WORK ENERGY AND POWER
355825
A block of mass 1 \(kg\) is pulled along the curved path \(ACB\) by a tangential force as shown in figure. The work done by the frictional force when the block moves from \(A\) to \(B\) is
1 5 \(J\)
2 10 \(J\)
3 20 \(J\)
4 45 \(J\)
Explanation:
Consider a small segment on the curve. Let \(\Delta S\) is the length of the segment Small work done by the external force is \(\begin{aligned}& \Delta W=F \Delta S \\& F=m g \sin \theta+\mu m g \cos \theta \\& \Delta W=m g \Delta S \sin \theta+\mu m g \Delta S \cos \theta \\& \Delta W=m g \Delta y+\mu m g \Delta x \\& W=\int_{y_{A}}^{y_{B}} m g \Delta y+\mu m g \int_{x_{A}}^{x_{A}} \Delta x \\& W=m g(0)+\mu m g\left(x_{B}-x_{A}\right) \\& W=\mu m g(10 m) \\& W=0.2 \times 1 \times 10(10)=20 \mathrm{~J}\end{aligned}\)
