A truck of mass \(30,000~\text{kg}\) moves up an inclined plane of slope \(1\) in \(100\) \(\left(\tan\theta = \frac{1}{100}\right)\) at a speed of \(30~\text{km/h}\). The power of the truck is: (given \(g=10~\text{ms}^{-2}\)):

1.	\(25~\text{kW}\)	2.	\(10~\text{kW}\)
3.	\(5~\text{kW}\)	4.	\(2.5~\text{kW}\)

A spring 40 mm long is stretched by the application of force. If 10 N force is required to stretch the spring through 1 mm, then work done to stretch the spring 40 mm is equal to:

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. \(m g \left(h + x \right) = \frac{1}{2} k \left(x + h \right)^{2}\)${\mathrm{}}^{}$

A block of mass \(M\) moving on the frictionless horizontal surface collides with the spring of spring constant \(k\) and compresses it by length \(L.\) The maximum momentum of the block after the collision will be:
                  

A body of mass 1 kg is thrown upwards with a velocity of $20$ ${\mathrm{ms}}^{-1}.$ It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? $\left(\mathrm{g}=10\right)$ ${\mathrm{ms}}^{-2}$

1. 20 J

2. 30 J

3. 40 J

4. 10 J