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.

A person of mass \(m\) ascends the stairs and goes up slowly through a height \(h\). Then,

A block is released from rest from a height of \(h = 5 ~\text m.\) After traveling through the smooth curved surface, it moves on the rough horizontal surface through a length \(l = 8 ~\text m\) and climbs onto the other smooth curved surface at a height \(h'.\) If $\mathrm{}$\(μ = 0.5,\) then the height \( h'\) is:

The potential energy of a particle of mass m varies as the magnitude of the $U=a{x}^2+by.$ The magnitude of the acceleration of the particle at (0, 3) is: (symbols have their usual meaning)

1.  $\sqrt{\frac{\mathrm{b}}{\mathrm{m}}}$

2.  $\sqrt{\frac{3\mathrm{b}}{\mathrm{m}}}$

3.  $\frac{\mathrm{b}}{\mathrm{m}}$

4.  Zero

A particle of mass \(10\) kg moves with a velocity of \(10\sqrt{x}\) in SI units, where \(x\) is displacement. The work done by the net force during the displacement of the particle from \(x=4~\text{m}\) to \(x= 9~\text{m}\) is:
1. \(1250~\text{J}\)
2. \(1000~\text{J}\)
3. \(3500~\text{J}\)
4. \(2500~\text{J}\)

A rigid body of mass \(m\) is moving in a circle of radius \(r\) with constant speed \(v.\) The force on the body is \(\dfrac{mv^2}{r}\) and is always directed towards the centre. The work done by this force in moving the body over half the circumference of the circle will be:
1. \(\dfrac{mv^{2}}{rπ}\) 
2. \(mr^{2} \pi\)
3. zero
4. \(2 mv^{2} \pi\)$\mathrm{}$

The kinetic energy of a body is increased by 21%. The percentage increase in the magnitude of linear momentum of the body will be:

1.  10%

2.  20%

3.  Zero

4.  11.5%

If a stone is projected vertically upward from the ground at a speed of 10 m/s, then it's: (g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

1.  Potential energy will be maximum after 0.5 s

2.  Kinetic energy will be maximum again after 1 s

3.  Kinetic energy = potential energy at a height of 2.5 m from the ground

4.  Potential energy will be minimum after 1 s