A force of 5 N making an angle $\theta$ with the horizontal acting on an object displaces it by 0.4 m along the horizontal direction. If the object gains kinetic energy of 1 J then the component of the force is:

1.	1.5 N	2.	2.5 N
3.	3.5 N	4.	4.5 N

A block of mass \(2~\text{kg}\) moving with a velocity of \(10~\text{m/s}\) on a smooth surface hits a spring of force constant \(80\times10^3~\text{N/m}\) as shown in the figure. The maximum compression in the spring will be:
               
1. \(5~\text{cm}\) 
2. \(10~\text{cm}\)
3. \(15~\text{cm}\)
4. \(20~\text{cm}\)

A block of mass \(m\) initially at rest, is dropped from a height \(h\) onto a spring of force constant \(k.\) If the maximum compression in the spring is \(x,\) then:     
            
1. \(m g h = \frac{1}{2} k x^{2}\)
2. \(m g \left(h + x\right) = \frac{1}{2} k x^{2}\)
3. \(m g h = \frac{1}{2} k \left(x + h\right)^{2}\)
4. \(m g \left(h + x \right) = \frac{1}{2} k \left(x + h \right)^{2}\)${\mathrm{}}^{}$

What is the work done by the gravitational force on block \(A\) during the first \(2\) s after the blocks are released? (assume the pulley is light)

1. \( 240 ~\text J\)
2. \( 200 ~\text J\)
3. \(120 ~\text J\)
4. \( 24 ~\text J\)

A weight 'mg' is suspended from a spring. The energy stored in the spring is U. The elongation in the spring is:

1.  $\frac{2\mathrm{U}}{\mathrm{mg}}$

2.  $\frac{\mathrm{U}}{\mathrm{mg}}$

3.  $\frac{\sqrt{2}\mathrm{U}}{\mathrm{mg}}$

4.  $\frac{\mathrm{U}}{\sqrt{2}\mathrm{mg}}$

      
1. \(50~\text{J}\)
2. \(100~\text{J}\)
3. \(25~\text{J}\)
4. Zero

A block is carried slowly up an inclined plane. If \(W_f\) is work done by the friction, \(W_N\)${\mathrm{}}_{}$ is work done by the reaction force, \(W_g\) is work done by the gravitational force, and \(W_{ex}\) is the work done by an external force, then choose the correct relation(s):
1. \(W _N +   W _f   +   W _g   +   W _{ex}   =   0\)
2. \(W _N   =   0\)
3. \(   W _f   +   W _{ex}   =    - W _g\)
4. All of these