355668
A coconut of mass \(m\) falls from the tree through a vertical distance of \(s\) and could reach ground with a velocity of \(vm{s^{ - 1}}\) due to air resistance. Work done by air resistance is
1 \(mgs\)
2 \(-\dfrac{1}{2} m v^{2}\)
3 \(m v^{2}+2 m g s\)
4 \(-\dfrac{m}{2}\left(2 g s-v^{2}\right)\)
Explanation:
Work done \(=\) Change in K.E. i.e., \(\quad W_{g}+W_{a i r}=\dfrac{1}{2} m v^{2}\) Where \(W_{g}=m g s\) and \(W_{a i r}\) is the work done by air resistance \(\begin{aligned}\therefore W_{a} & =-m g s+\dfrac{1}{2} m v^{2} \\& =-\dfrac{m}{2}\left(-v^{2}+2 g s\right)=-\dfrac{m}{2}\left(2 g s-v^{2}\right)\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355669
A force \(F\), making an angle \(\theta\) with the horizontal, acting on an object displaces it by 0.4 \(m\) along the horizontal direction. If the object gains kinetic energy of 1 \(J\) the horizontal component of the force is
1 2.5 \(N\)
2 1.5 \(N\)
3 4.5 \(N\)
4 3.5 \(N\)
Explanation:
Work done on the body\(=K.E.\) gained by the body \(F s \cos \theta=1 \Rightarrow F \cos \theta=\dfrac{1}{s}=\dfrac{1}{0.4}=2.5 N \text {. }\)
PHXI06:WORK ENERGY AND POWER
355670
A body of mass \(M\) is dropped from a height \(h\) on a sand floor. If the body penetrates \(x \)\(cm\) into the sand, the average resistance offered by the sand to the body is
1 \(M g\left(1-\dfrac{h}{x}\right)\)
2 \(Mg\left( {\frac{h}{x}} \right)\)
3 \(M g h+m g x\)
4 \(M g\left(1+\dfrac{h}{x}\right)\)
Explanation:
When body passes through the sand floor it comes to rest after travelling a distance \(\mathrm{x}\). Let \(F\) be the resisting force acting on the body. Work done by all the forces is equal to change in \(\mathrm{KE}\) \(\mathrm{Fx}=\mathrm{Mgh}+\operatorname{Mgx} \text { (F-resistive force) }\) \(F=M g\left(1+\dfrac{h}{x}\right)\)
PHXI06:WORK ENERGY AND POWER
355671
A body starts from rest and acquires a velocity \(V\) in time \(T\). The work done on the body in time \(t\) will be proportional to :
1 \(\dfrac{V^{2} t^{2}}{T}\)
2 \(\dfrac{V}{T} t\)
3 \(\dfrac{V^{2}}{T^{2}} t^{2}\)
4 \(\dfrac{V^{2}}{T^{2}} t\)
Explanation:
Acceleration of the body : \(a=V / T\). Velocity acquired in time \(t\) \(v=a t \Rightarrow v=\dfrac{V}{T} t\) \(\Rightarrow \quad K \cdot E \propto v^{2}\) Work done on the body is equal to gain in the kinetic energy. So Work done \(\alpha \dfrac{V^{2} t^{2}}{T^{2}}\)
PHXI06:WORK ENERGY AND POWER
355672
A block of mass 5 \(kg\) is resting on a smooth surface. At what angle a force of 20 \(N\) be acted on the body so that it will acquire a kinetic energy of 40 \(J\) after moving 4 \(m\) ?
1 \(120^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
According to work-energy theorem \(W=\) Change in kinetic energy \(F S \cos \theta=\dfrac{1}{2} m v^{2}-\dfrac{1}{2} m u^{2}\) Substituting the given values, we get \(\begin{aligned}& 20 \times 4 \times \cos \theta=40-0 \\& \cos \theta=\dfrac{40}{80}=\dfrac{1}{2} \\& \theta=\cos ^{-1}\left(\dfrac{1}{2}\right)=60^{\circ}\end{aligned}\)
355668
A coconut of mass \(m\) falls from the tree through a vertical distance of \(s\) and could reach ground with a velocity of \(vm{s^{ - 1}}\) due to air resistance. Work done by air resistance is
1 \(mgs\)
2 \(-\dfrac{1}{2} m v^{2}\)
3 \(m v^{2}+2 m g s\)
4 \(-\dfrac{m}{2}\left(2 g s-v^{2}\right)\)
Explanation:
Work done \(=\) Change in K.E. i.e., \(\quad W_{g}+W_{a i r}=\dfrac{1}{2} m v^{2}\) Where \(W_{g}=m g s\) and \(W_{a i r}\) is the work done by air resistance \(\begin{aligned}\therefore W_{a} & =-m g s+\dfrac{1}{2} m v^{2} \\& =-\dfrac{m}{2}\left(-v^{2}+2 g s\right)=-\dfrac{m}{2}\left(2 g s-v^{2}\right)\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355669
A force \(F\), making an angle \(\theta\) with the horizontal, acting on an object displaces it by 0.4 \(m\) along the horizontal direction. If the object gains kinetic energy of 1 \(J\) the horizontal component of the force is
1 2.5 \(N\)
2 1.5 \(N\)
3 4.5 \(N\)
4 3.5 \(N\)
Explanation:
Work done on the body\(=K.E.\) gained by the body \(F s \cos \theta=1 \Rightarrow F \cos \theta=\dfrac{1}{s}=\dfrac{1}{0.4}=2.5 N \text {. }\)
PHXI06:WORK ENERGY AND POWER
355670
A body of mass \(M\) is dropped from a height \(h\) on a sand floor. If the body penetrates \(x \)\(cm\) into the sand, the average resistance offered by the sand to the body is
1 \(M g\left(1-\dfrac{h}{x}\right)\)
2 \(Mg\left( {\frac{h}{x}} \right)\)
3 \(M g h+m g x\)
4 \(M g\left(1+\dfrac{h}{x}\right)\)
Explanation:
When body passes through the sand floor it comes to rest after travelling a distance \(\mathrm{x}\). Let \(F\) be the resisting force acting on the body. Work done by all the forces is equal to change in \(\mathrm{KE}\) \(\mathrm{Fx}=\mathrm{Mgh}+\operatorname{Mgx} \text { (F-resistive force) }\) \(F=M g\left(1+\dfrac{h}{x}\right)\)
PHXI06:WORK ENERGY AND POWER
355671
A body starts from rest and acquires a velocity \(V\) in time \(T\). The work done on the body in time \(t\) will be proportional to :
1 \(\dfrac{V^{2} t^{2}}{T}\)
2 \(\dfrac{V}{T} t\)
3 \(\dfrac{V^{2}}{T^{2}} t^{2}\)
4 \(\dfrac{V^{2}}{T^{2}} t\)
Explanation:
Acceleration of the body : \(a=V / T\). Velocity acquired in time \(t\) \(v=a t \Rightarrow v=\dfrac{V}{T} t\) \(\Rightarrow \quad K \cdot E \propto v^{2}\) Work done on the body is equal to gain in the kinetic energy. So Work done \(\alpha \dfrac{V^{2} t^{2}}{T^{2}}\)
PHXI06:WORK ENERGY AND POWER
355672
A block of mass 5 \(kg\) is resting on a smooth surface. At what angle a force of 20 \(N\) be acted on the body so that it will acquire a kinetic energy of 40 \(J\) after moving 4 \(m\) ?
1 \(120^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
According to work-energy theorem \(W=\) Change in kinetic energy \(F S \cos \theta=\dfrac{1}{2} m v^{2}-\dfrac{1}{2} m u^{2}\) Substituting the given values, we get \(\begin{aligned}& 20 \times 4 \times \cos \theta=40-0 \\& \cos \theta=\dfrac{40}{80}=\dfrac{1}{2} \\& \theta=\cos ^{-1}\left(\dfrac{1}{2}\right)=60^{\circ}\end{aligned}\)
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PHXI06:WORK ENERGY AND POWER
355668
A coconut of mass \(m\) falls from the tree through a vertical distance of \(s\) and could reach ground with a velocity of \(vm{s^{ - 1}}\) due to air resistance. Work done by air resistance is
1 \(mgs\)
2 \(-\dfrac{1}{2} m v^{2}\)
3 \(m v^{2}+2 m g s\)
4 \(-\dfrac{m}{2}\left(2 g s-v^{2}\right)\)
Explanation:
Work done \(=\) Change in K.E. i.e., \(\quad W_{g}+W_{a i r}=\dfrac{1}{2} m v^{2}\) Where \(W_{g}=m g s\) and \(W_{a i r}\) is the work done by air resistance \(\begin{aligned}\therefore W_{a} & =-m g s+\dfrac{1}{2} m v^{2} \\& =-\dfrac{m}{2}\left(-v^{2}+2 g s\right)=-\dfrac{m}{2}\left(2 g s-v^{2}\right)\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355669
A force \(F\), making an angle \(\theta\) with the horizontal, acting on an object displaces it by 0.4 \(m\) along the horizontal direction. If the object gains kinetic energy of 1 \(J\) the horizontal component of the force is
1 2.5 \(N\)
2 1.5 \(N\)
3 4.5 \(N\)
4 3.5 \(N\)
Explanation:
Work done on the body\(=K.E.\) gained by the body \(F s \cos \theta=1 \Rightarrow F \cos \theta=\dfrac{1}{s}=\dfrac{1}{0.4}=2.5 N \text {. }\)
PHXI06:WORK ENERGY AND POWER
355670
A body of mass \(M\) is dropped from a height \(h\) on a sand floor. If the body penetrates \(x \)\(cm\) into the sand, the average resistance offered by the sand to the body is
1 \(M g\left(1-\dfrac{h}{x}\right)\)
2 \(Mg\left( {\frac{h}{x}} \right)\)
3 \(M g h+m g x\)
4 \(M g\left(1+\dfrac{h}{x}\right)\)
Explanation:
When body passes through the sand floor it comes to rest after travelling a distance \(\mathrm{x}\). Let \(F\) be the resisting force acting on the body. Work done by all the forces is equal to change in \(\mathrm{KE}\) \(\mathrm{Fx}=\mathrm{Mgh}+\operatorname{Mgx} \text { (F-resistive force) }\) \(F=M g\left(1+\dfrac{h}{x}\right)\)
PHXI06:WORK ENERGY AND POWER
355671
A body starts from rest and acquires a velocity \(V\) in time \(T\). The work done on the body in time \(t\) will be proportional to :
1 \(\dfrac{V^{2} t^{2}}{T}\)
2 \(\dfrac{V}{T} t\)
3 \(\dfrac{V^{2}}{T^{2}} t^{2}\)
4 \(\dfrac{V^{2}}{T^{2}} t\)
Explanation:
Acceleration of the body : \(a=V / T\). Velocity acquired in time \(t\) \(v=a t \Rightarrow v=\dfrac{V}{T} t\) \(\Rightarrow \quad K \cdot E \propto v^{2}\) Work done on the body is equal to gain in the kinetic energy. So Work done \(\alpha \dfrac{V^{2} t^{2}}{T^{2}}\)
PHXI06:WORK ENERGY AND POWER
355672
A block of mass 5 \(kg\) is resting on a smooth surface. At what angle a force of 20 \(N\) be acted on the body so that it will acquire a kinetic energy of 40 \(J\) after moving 4 \(m\) ?
1 \(120^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
According to work-energy theorem \(W=\) Change in kinetic energy \(F S \cos \theta=\dfrac{1}{2} m v^{2}-\dfrac{1}{2} m u^{2}\) Substituting the given values, we get \(\begin{aligned}& 20 \times 4 \times \cos \theta=40-0 \\& \cos \theta=\dfrac{40}{80}=\dfrac{1}{2} \\& \theta=\cos ^{-1}\left(\dfrac{1}{2}\right)=60^{\circ}\end{aligned}\)
355668
A coconut of mass \(m\) falls from the tree through a vertical distance of \(s\) and could reach ground with a velocity of \(vm{s^{ - 1}}\) due to air resistance. Work done by air resistance is
1 \(mgs\)
2 \(-\dfrac{1}{2} m v^{2}\)
3 \(m v^{2}+2 m g s\)
4 \(-\dfrac{m}{2}\left(2 g s-v^{2}\right)\)
Explanation:
Work done \(=\) Change in K.E. i.e., \(\quad W_{g}+W_{a i r}=\dfrac{1}{2} m v^{2}\) Where \(W_{g}=m g s\) and \(W_{a i r}\) is the work done by air resistance \(\begin{aligned}\therefore W_{a} & =-m g s+\dfrac{1}{2} m v^{2} \\& =-\dfrac{m}{2}\left(-v^{2}+2 g s\right)=-\dfrac{m}{2}\left(2 g s-v^{2}\right)\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355669
A force \(F\), making an angle \(\theta\) with the horizontal, acting on an object displaces it by 0.4 \(m\) along the horizontal direction. If the object gains kinetic energy of 1 \(J\) the horizontal component of the force is
1 2.5 \(N\)
2 1.5 \(N\)
3 4.5 \(N\)
4 3.5 \(N\)
Explanation:
Work done on the body\(=K.E.\) gained by the body \(F s \cos \theta=1 \Rightarrow F \cos \theta=\dfrac{1}{s}=\dfrac{1}{0.4}=2.5 N \text {. }\)
PHXI06:WORK ENERGY AND POWER
355670
A body of mass \(M\) is dropped from a height \(h\) on a sand floor. If the body penetrates \(x \)\(cm\) into the sand, the average resistance offered by the sand to the body is
1 \(M g\left(1-\dfrac{h}{x}\right)\)
2 \(Mg\left( {\frac{h}{x}} \right)\)
3 \(M g h+m g x\)
4 \(M g\left(1+\dfrac{h}{x}\right)\)
Explanation:
When body passes through the sand floor it comes to rest after travelling a distance \(\mathrm{x}\). Let \(F\) be the resisting force acting on the body. Work done by all the forces is equal to change in \(\mathrm{KE}\) \(\mathrm{Fx}=\mathrm{Mgh}+\operatorname{Mgx} \text { (F-resistive force) }\) \(F=M g\left(1+\dfrac{h}{x}\right)\)
PHXI06:WORK ENERGY AND POWER
355671
A body starts from rest and acquires a velocity \(V\) in time \(T\). The work done on the body in time \(t\) will be proportional to :
1 \(\dfrac{V^{2} t^{2}}{T}\)
2 \(\dfrac{V}{T} t\)
3 \(\dfrac{V^{2}}{T^{2}} t^{2}\)
4 \(\dfrac{V^{2}}{T^{2}} t\)
Explanation:
Acceleration of the body : \(a=V / T\). Velocity acquired in time \(t\) \(v=a t \Rightarrow v=\dfrac{V}{T} t\) \(\Rightarrow \quad K \cdot E \propto v^{2}\) Work done on the body is equal to gain in the kinetic energy. So Work done \(\alpha \dfrac{V^{2} t^{2}}{T^{2}}\)
PHXI06:WORK ENERGY AND POWER
355672
A block of mass 5 \(kg\) is resting on a smooth surface. At what angle a force of 20 \(N\) be acted on the body so that it will acquire a kinetic energy of 40 \(J\) after moving 4 \(m\) ?
1 \(120^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
According to work-energy theorem \(W=\) Change in kinetic energy \(F S \cos \theta=\dfrac{1}{2} m v^{2}-\dfrac{1}{2} m u^{2}\) Substituting the given values, we get \(\begin{aligned}& 20 \times 4 \times \cos \theta=40-0 \\& \cos \theta=\dfrac{40}{80}=\dfrac{1}{2} \\& \theta=\cos ^{-1}\left(\dfrac{1}{2}\right)=60^{\circ}\end{aligned}\)
355668
A coconut of mass \(m\) falls from the tree through a vertical distance of \(s\) and could reach ground with a velocity of \(vm{s^{ - 1}}\) due to air resistance. Work done by air resistance is
1 \(mgs\)
2 \(-\dfrac{1}{2} m v^{2}\)
3 \(m v^{2}+2 m g s\)
4 \(-\dfrac{m}{2}\left(2 g s-v^{2}\right)\)
Explanation:
Work done \(=\) Change in K.E. i.e., \(\quad W_{g}+W_{a i r}=\dfrac{1}{2} m v^{2}\) Where \(W_{g}=m g s\) and \(W_{a i r}\) is the work done by air resistance \(\begin{aligned}\therefore W_{a} & =-m g s+\dfrac{1}{2} m v^{2} \\& =-\dfrac{m}{2}\left(-v^{2}+2 g s\right)=-\dfrac{m}{2}\left(2 g s-v^{2}\right)\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355669
A force \(F\), making an angle \(\theta\) with the horizontal, acting on an object displaces it by 0.4 \(m\) along the horizontal direction. If the object gains kinetic energy of 1 \(J\) the horizontal component of the force is
1 2.5 \(N\)
2 1.5 \(N\)
3 4.5 \(N\)
4 3.5 \(N\)
Explanation:
Work done on the body\(=K.E.\) gained by the body \(F s \cos \theta=1 \Rightarrow F \cos \theta=\dfrac{1}{s}=\dfrac{1}{0.4}=2.5 N \text {. }\)
PHXI06:WORK ENERGY AND POWER
355670
A body of mass \(M\) is dropped from a height \(h\) on a sand floor. If the body penetrates \(x \)\(cm\) into the sand, the average resistance offered by the sand to the body is
1 \(M g\left(1-\dfrac{h}{x}\right)\)
2 \(Mg\left( {\frac{h}{x}} \right)\)
3 \(M g h+m g x\)
4 \(M g\left(1+\dfrac{h}{x}\right)\)
Explanation:
When body passes through the sand floor it comes to rest after travelling a distance \(\mathrm{x}\). Let \(F\) be the resisting force acting on the body. Work done by all the forces is equal to change in \(\mathrm{KE}\) \(\mathrm{Fx}=\mathrm{Mgh}+\operatorname{Mgx} \text { (F-resistive force) }\) \(F=M g\left(1+\dfrac{h}{x}\right)\)
PHXI06:WORK ENERGY AND POWER
355671
A body starts from rest and acquires a velocity \(V\) in time \(T\). The work done on the body in time \(t\) will be proportional to :
1 \(\dfrac{V^{2} t^{2}}{T}\)
2 \(\dfrac{V}{T} t\)
3 \(\dfrac{V^{2}}{T^{2}} t^{2}\)
4 \(\dfrac{V^{2}}{T^{2}} t\)
Explanation:
Acceleration of the body : \(a=V / T\). Velocity acquired in time \(t\) \(v=a t \Rightarrow v=\dfrac{V}{T} t\) \(\Rightarrow \quad K \cdot E \propto v^{2}\) Work done on the body is equal to gain in the kinetic energy. So Work done \(\alpha \dfrac{V^{2} t^{2}}{T^{2}}\)
PHXI06:WORK ENERGY AND POWER
355672
A block of mass 5 \(kg\) is resting on a smooth surface. At what angle a force of 20 \(N\) be acted on the body so that it will acquire a kinetic energy of 40 \(J\) after moving 4 \(m\) ?
1 \(120^{\circ}\)
2 \(30^{\circ}\)
3 \(45^{\circ}\)
4 \(60^{\circ}\)
Explanation:
According to work-energy theorem \(W=\) Change in kinetic energy \(F S \cos \theta=\dfrac{1}{2} m v^{2}-\dfrac{1}{2} m u^{2}\) Substituting the given values, we get \(\begin{aligned}& 20 \times 4 \times \cos \theta=40-0 \\& \cos \theta=\dfrac{40}{80}=\dfrac{1}{2} \\& \theta=\cos ^{-1}\left(\dfrac{1}{2}\right)=60^{\circ}\end{aligned}\)