Q. No.Given below are two statements:
Assertion (A):
Work done by friction is always negative.
Reason (R):
Kinetic friction is a non-conservative force.
1.
Both (A) and (R) are True and (R) is the correct explanation of (A).
2.
Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.
(A) is True but (R) is False.
4.
(A) is False but (R) is True.
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A block of mass \(m\) is being lowered by means of a string attached to it. The system moves down with a constant velocity. Then:
| 1. | the work done by gravity on the block is positive. |
| 2. | the work done by force, \(F \) (the force of the string) on the block is negative. |
| 3. | the work done by gravity is equal in magnitude to that done by the string. |
| 4. | All of the above are true. |
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A block of mass \(m\) is placed on a vertical spring and compressed downward by an external force. When released, the block moves upward. Let the work done by the spring be \(W_s,\) and the work done by gravity be \(W_g.\) Denote the potential energy stored in the spring as \(P_s,\) the gravitational potential energy of the block as \(P_{mg},\) and its kinetic energy as \(K_m.\)
Which of the following relations correctly represents the energy transformations during the block’s motion?
| 1. | loss in \(P_s\) \(=\) gain in \(K_m\) |
| 2. | loss in \(P_s\) \(=\) gain in \(K_m\) \(+\) gain in \(P_{mg}\) |
| 3. | \(W_g = W_s \) |
| 4. | \(W_s\) \(=\) gain in \(K_m\) |
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The potential energy due to a force is given by:
\(U(x,y)= -3xy+2y^2\) (in joule)
where \(x,y\) are in metres.
The force acting when \(x=0,y=1\) (m) is: (in magnitude)
1. \(2~\text{N}\)
2. \(1~\text{N}\)
3. \(3~\text{N}\)
4. \(5~\text{N}\)
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The force acting on a particle is shown in the diagram as a function of \(x\). Work done by this force when the particle moves from \(x=0~\text{to}~x=2~\text{m}\) equals:
| 1. | \(5~\text{J}\) | 2. | \(10~\text{J}\) |
| 3. | \(7.5~\text{J}\) | 4. | \(2.5~\text{J}\) |
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A simple pendulum consisting of a bob of mass \(m\), and a string of length \(L\) is given a horizontal speed \(u\), at its lowest point as shown in the figure. As a result, it rises to \(B\), where it just comes to rest momentarily with \(OB\) horizontal.
During the motion \(AB,\)
| 1. | Work done by the string is zero |
| 2. | Work done by gravity is \(-mgL\) |
| 3. | Change in K.E. of the bob is \(-\dfrac{1}{2}mu^2\) |
| 4. | All the above are true |
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A person of mass \(m\) ascends the stairs and goes up slowly through a height \(h\). Then,
| 1. | Work done by gravity is \(mgh\) |
| 2. | Work done by normal reaction is \(mgh\) |
| 3. | Work done by normal reaction is zero |
| 4. | Work done by gravity is stored as gravitational \(P.E\). |
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A particle of mass '\(m\)' is released from the origin, and it moves under the action of a force: \(F(x)= F_0-kx\)
The maximum speed of the particle is, \(v= \)
| 1. | \(\sqrt{\dfrac{F_0^2}{mk}}\) | 2. | \(\sqrt{\dfrac{2F_0^2}{mk}}\) |
| 3. | \(\sqrt{\dfrac{F_0^2}{2mk}}\) | 4. | \(2\sqrt{\dfrac{F_0^2}{mk}}\) |
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A projectile is launched from a cliff of height \(h,\) with an initial speed \(u,\) at an angle \(\theta.\) The speed with which it hits the ground:
| 1. | depends on the vertical component, \(u \text{sin}\theta\) |
| 2. | depends on the horizontal component, \(u \text{cos}\theta\) |
| 3. | depends on \(u,\) but not on \(\theta\) |
| 4. | depends on the quantity \(u \text{tan}\theta\) |
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A small block of mass '\(m\)' is placed against a compressed spring, of spring constant \(k\). The initial compression in the spring is '\(d\)'. The block is released and the spring relaxes, while the block is projected up to a height \(H\) relative to its initial position. Then, \(H\) =
| 1. | \(\dfrac{kd^2}{2mg}\) | 2. | \(\dfrac{kd^2}{2mg}+d \) |
| 3. | \(\dfrac{kd^2}{2mg}-d\) | 4. | \(\dfrac{kd^2}{mg}+d\) |
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