# Two equal masses, $$m_1$$ and $$m_2,$$ moving in the same straight line at velocities +3 m/s and –5 m/s respectively, collide elastically. Their velocities after the collision will be: 1. +4 m/s for both 2. –3 m/s and +5 m/s 3. –4 m/s and +4 m/s 4. –5 m/s and +3 m/s

Subtopic:  Collisions |
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A uniform chain of length $$L$$ and mass $$M$$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $$g$$ is acceleration due to gravity, the work required to pull the hanging part on the table is:

1. $$MgL$$

2. $$MgL/3$$

3. $$MgL/9$$

4. $$MgL/18$$

Subtopic:  Gravitational Potential Energy |
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A particle of mass 'm' is moving in a horizontal circle of radius 'r' under a centripetal force equal to –K/r2, where K is a constant. The total energy of the particle will be:

1. $\frac{K}{2r}$

2. $-\frac{K}{2r}$

3. $-\frac{K}{r}$

4. $\frac{K}{r}$

Subtopic:  Potential Energy: Relation with Force |
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A force $\mathbit{F}=-k\left(y\mathbit{i}+x\mathbit{j}\right)$ (where k is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is:

1. $-2k{a}^{2}$

2. $2k{a}^{2}$

3. $-k{a}^{2}$

4. $$ka^2$$

Subtopic:  Work Done by Variable Force |
56%
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A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force, then:

1. Lorry will come to rest in a shorter distance

2. Car will come to rest in a shorter distance

3. Both will come to rest in a same distance

4. None of the above

Subtopic:  Work Energy Theorem |
64%
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The relationship between force and position is shown in the given figure (in a one-dimensional case). The work done by the force in displacing a body from $$x = 1$$ cm to $$x = 5$$ cm is:

1. $$20$$ ergs
2. $$60$$ ergs
3. $$70$$ ergs
4. $$700$$ ergs

Subtopic:  Work Done by Variable Force |
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The graph between the resistive force $$F$$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $$25$$ kg and the initial velocity is $$2$$ m/s. When the distance covered by the body is $$4$$ m, its kinetic energy would be:

1. $$50$$ J
2. $$40$$ J
3. $$20$$ J
4. $$10$$ J

Subtopic:  Work Done by Variable Force |
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The relationship between the force F and the position x of a body is as shown in the figure. The work done in displacing the body from x = 1 m to x = 5 m will be:

 1 30 J 2 15 J 3 25 J 4 20 J
Subtopic:  Work Done by Variable Force |
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A particle moves from a point $\left(-2\stackrel{^}{i}+5\stackrel{^}{j}\right)$ to $\left(4\stackrel{^}{j}+3\stackrel{^}{k}\right)$ when a force of  $\left(4\stackrel{^}{i}+3\stackrel{^}{j}\right)$ N is applied. How much work has been done by the force?

 1 8 J 2 11 J 3 5 J 4 2 J
Subtopic:  Work done by constant force |
68%
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NEET - 2016
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A block of mass $$10$$ kg, moving in the $$x\text-$$direction with a constant speed of $$10$$ ms-1, is subjected to a retarding force $$F=0.1x$$ J/m during its travel from $$x =20$$ m to $$30$$ m. Its final kinetic energy will be:

 1 $$475$$ J 2 $$450$$ J 3 $$275$$ J 4 $$250$$ J
Subtopic:  Work Done by Variable Force |
72%
From NCERT
NEET - 2015
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