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A block is connected to a wall by an elastic spring, which is initially unextended. A constant force \(F\) is applied to the block, so that the spring is stretched and the block moves to the right. The spring extends and slows the block down, and it comes to rest again, briefly. There is no friction anywhere. The displacement of the block is \(d.\)
           
The work done by the force \(F\) equals:
1. \(\dfrac{Fd}{2}\)
2. \(Fd\)
3. \(2Fd\)
4. \(\dfrac{Fd}{3}\)

Subtopic:  Work done by constant force |
 58%
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A block is connected to a wall by an elastic spring, which is initially unextended. A constant force \(F\) is applied to the block, so that the spring is stretched and the block moves to the right. The spring extends and slows the block down, and it comes to rest again, briefly. There is no friction anywhere. The displacement of the block is \(d.\)
           
The potential energy stored in the spring equals:
1. \(\Large\frac{Fd}{2}\) 2. \(Fd\)
3. \(\Large\frac{2Fd}{3}\) 4. \(\Large\frac{Fd}{3}\)
Subtopic:  Elastic Potential Energy |
 57%
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A block is connected to a wall by an elastic spring, which is initially unextended. A constant force \(F\) is applied to the block, so that the spring is stretched and the block moves to the right. The spring extends and slows the block down, and it comes to rest again, briefly. There is no friction anywhere. The displacement of the block is \(d.\)
           
The force exerted by the spring equals:
1. \(F\)
2. \(\Large\frac F2\)
3. \(2F\)
4. \(\sqrt2F\)
Subtopic:  Work Energy Theorem |
 53%
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What is the minimum speed that must be imparted to the bob of a pendulum (at \(A\)), so that it rises to the level of \(O,\) during subsequent motion? The initial position \(A\) and the dimensions are indicated in the figure.
1. \(\sqrt{gL}\) 2. \(\sqrt{\Large\frac{gL}{2}}\)
3. \(\sqrt{2gL}\) 4. \(\sqrt{\Large\frac{5gL}{2}}\)
Subtopic:  Conservation of Mechanical Energy |
 56%
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Given below are two statements: 
Assertion (A): Work done by all forces, conservative or non-conservative, equals the change in the total energy of the system.
Reason (R): According to Newton's laws of motion, the acceleration of any particle of the system is due to all forces acting on it; and therefore the work done on it by these is the total work done.
 
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.
Subtopic:  Work Energy Theorem |
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A block of mass \(m\) collides with another of mass \(3m,\) which is at rest, initially. After the collision, the first block comes to rest. The coefficient of restitution (i.e, the ratio of the relative speed after the collision to the relative speed before the collision.) is:
1. \(\dfrac{1}{2}\) 2. \(\dfrac{1}{3}\)
3. \(\dfrac{1}{9}\) 4. \(\dfrac{1}{\sqrt3}\)
Subtopic:  Collisions |
 83%
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A man (mass : \(m\)) climbs a staircase of height \(h,\) slowly. The work done on the man by the reaction force exerted by the stairs is:
1. \(mgh\)
2. \(\Large\frac{mgh}{2}\)
3. \(-mgh\)
4. zero
Subtopic:  Concept of Work |
 58%
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An elastic ball, released from a height, loses \(50\%\) of its kinetic energy on impact with the ground, which is horizontal. The coefficient of restitution is:
1. \({\Large\frac12}\)
2. \({\Large\frac{1}{\sqrt2}}\)
3. \({\Large\frac14}\)
4. \(\sqrt2-1\)
Subtopic:  Collisions |
 81%
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Two blocks of masses \(m,2m\) are held against two compressed springs and released on a smooth, horizontal surface. After losing contact with the springs, they collide and come to rest. The energies of the springs before the blocks are released are \(E_1\) and \(E_2,\) while the masses of the blocks are \(m\) and \(2m,\) as shown in the figure. Then,
                             
1. \(E_1=E_2\)
2. \(E_1=2E_2\)
3. \(2E_1=E_2\)
4. \(E_1=4E_2\)
Subtopic:  Elastic Potential Energy |
 66%
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A \(200~\text g\) ball is thrown upward by applying constant upward force \((F)\) over a distance of \(50~\text{cm}\) and then releasing it;    the ball ascends to a height of \(10~\text m\) after it is released. The work done by the applied force \(F\) is: (take \(g=10~\text{m/s}^2\))
1. \(20~\text J\)
2. \(19~\text J\)
3. \(21~\text J\)
4. \(42~\text J\)
Subtopic:  Work done by constant force |
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