The body of a man \(100~\text{kg}\) travelled \(10~\text{m}\) before coming to rest. If \(\mu=0.4,\) then work done against friction is: (motion is happening on a horizontal surface, take \(g=10~\text{m/s}^2\))
1. \(4500~\text{J}\)
2. \(5000~\text{J}\)
3. \(4200~\text{J}\)
4. \(4000~\text{J}\)
Subtopic:  Work Energy Theorem |
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It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass \(1.00\) g falling from a height of \(1.00\) km. It hits the ground with a speed of \(50.0\) m/s. Work done by the gravitational force and work done by the unknown resistive force respectively are:

1. \(-8.75\) J and \(10\) J 2. \(10\) J and \(-8.75\) J
3. \(0\) J and \(2.26\) J 4. \(-10\) J and \(-10\) J
Subtopic:  Work Energy Theorem |
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If a number of forces act on the body and the body is in static or dynamic equilibrium, then:
1. work done by individual forces must be zero
2. net work done is positive
3. net work done is negative
4. net work done is zero
Subtopic:  Work Energy Theorem |
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A body is moved quasi-statically from point \(A\) to point \(C\) in a uniform gravitational field. During the motion, a constant vertical upward force \(F\) of magnitude \(mg\) is applied to keep the body in equilibrium. The body is moved either directly along the inclined path \(AC\) or along the path \(A \rightarrow B\rightarrow C,\) as shown in the figure. If \(W_{AC}\)​ and \(W_{ABC}\)​ denote the work done by the applied force \(F\) along the two paths, respectively, then:

1. \(W_{AC}=W_{ABC}\)
2. \(W_{AC}>W_{ABC}\)
3. \(W_{AC}<W_{ABC}\)
4. None of the above

Subtopic:  Work Energy Theorem |
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A \(500 ~\text{kg}\) object slides across a horizontal surface where the coefficient of kinetic friction is \(\mu=0.7.\) To maintain a constant velocity of \(10~\text{m/s},\) an external force is applied horizontally in the direction of motion. The object travels a total distance of \(4~\text{km}\) under these conditions. The work done by the external force is:
1. \(3.5 \times 10^6~\text{J} \)
2. \(28 \times 10^6~\text{J} \)
3. \(7 \times 10^6 ~\text{J} \)
4. \(14 \times 10^6 ~\text{J} \)

Subtopic:  Work Energy Theorem |
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\(300\) J of work is done in sliding a \(2\) kg block up an inclined plane of height \(10\) m. Taking \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), work done against friction is:
1. \(1000\) J
2. \(200\) J
3. \(100\) J
4. zero
Subtopic:  Work Energy Theorem |
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A \(2\) kg particle moves along \(x\)-axis such that its position \((x)\) varies with time \((t)\) as \(x = 2t^{2}+3. \) During the initial \(5\) s, the work done by all the forces acting on the particle is:
1. \(400\) J
2. \(500\) J
3. \(600\) J
4. \(900\) J

Subtopic:  Work Energy Theorem |
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A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled \(3~\text{m}\) is:

      

1. \(2.5~\text{J}\)
2. \(6.5~\text{J}\)
3. \(4~\text{J}\)
4. \(5~\text{J}\)

Subtopic:  Work Energy Theorem |
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A particle experiences a variable force \(\vec F = (4x \hat i + 3y^2 \hat j)\) in a horizontal \(x\text{-}y\) plane. Assume distance in meters and force is in newton. If the particle moves from point \((1,2)\) to point \((2,3)\) in the \(x\text{-}y\) plane, the kinetic energy changes by:
1. \(50.0\) J
2. \(12.5\) J
3. \(25.0\) J
4. \(0\) 
Subtopic:  Work Energy Theorem |
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A constant force acts on a body of mass \(1\) kg, providing it a kinetic energy of \(1800~\text{J}\) by the end of \(5^{\mathrm{th}}\) second. If the body was initially at rest, at the beginning of the action of force, then the magnitude of the force is equal to:

              
1. \(30~\text{ N}\)
2. \(12~\text{ N}\)
3. \(25~\text{ N}\)
4. \(15~\text{ N}\)
Subtopic:  Work Energy Theorem |
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