Question

A 100 kg lock is started with a speed of 2.0 m s−1 on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. (b) Consider the situation from a frame of reference moving at 2.0 m s−1 along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m s−1. calculate the increase in the kinetic energy of the block as it stops slipping past the belt. (c) Find the work done in this frame by the external force holding the belt.

Answer 1

(a)Given,

Mass of block , m = 100kg

Initial velocity = 2m/s

Coefficient of kinetic friction between

the block and the belt , u = 0.20

change in internal energy of the block

belt system = work done by the

internal force(friction

in this case)

= u*normal reaction

= 0.20*mg

= 0.20*1000

= 200J

(b) Given,

velocity of the given frame of

reference = 2m/s

relative initial velocity of block = initial

velocity of block w.r.t ground -velocity

of frame of reference = (2-2) = 0 m/s

relative final velocity of block = final

velocity of block w.r.t ground -

velocity of frame of reference = (0-2)

= -2 m/s

So, increase in kinetic energy w. r. t

given frame of reference= (final –

initial) kinetic energy = 1/2*100*[(-2) ^2

- (0) ^2] = 200J

(c) acceleration of block = frictional

force/ mass of block = (u*mg) /100=

(200/100) = 2 m/s^2

distance travelled by the block w. r. t

given frame of reference = (V^2 - U^2)

/2a = (4 - 0) /4 = 1 m

So, work done by friction = frictional f

force*distance travelled by the block

= 200*1 = 200J

work done to give relative final

velocity = change in relative kinetic

energy = 200J

Total work done by the external force

holding the belt w. r. t given frame of

reference = work done by the friction

+ work done to increase the kinetic

energy = 200 + 200 = 400J

Answer 2

(a) The change in the internal energy of the block-belt system as the block comes to a stop on the belt is 200 J

(b) The increase in the kinetic energy of the block as it stops slipping past the belt is 200 J.

(c) The work done in this frame by the external force holding the belt is 400J

Explanation:

Given

m = 100 kg

u = 2.0 m/s

v = 0

μk = 0.2

a) To calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt :

The belt-block system's internal energy will reduce when the block loses its Kinetic energy in heat due to friction. Accordingly,

$\text { KE lost }=\frac{1}{2} m u^{2}-\frac{1}{2} m v^{2}$

              $=\frac{1}{2} m\left(u^{2}-v^{2}\right)$

              $=\frac{1}{2} \times 100 \times\left(2^{2}-0^{2}\right)$

              $=50 \times\left(2^{2}\right)$

              $=50 \times 4$

              = 200 J  

b) To calculate the increase in the kinetic energy of the block as it stops slipping past the belt :

The frame velocity is given by the

$u_{\ell}=2.0 \mathrm{m} / \mathrm{s}$

$u^{\prime}=u-u_{f}=2-2=0$

$y^{\prime}=0-2=-2 \mathrm{m} / \mathrm{s}$

Kinetic energy lost = $\frac{1}{2} m u^{\prime 2}-\frac{1}{2} m v^{\prime 2}$

$=\frac{1}{2} m\left(u^{2}-v^{2}\right)$

$=\frac{1}{2} \times 100 \times\left(0^{2}-2^{2}\right)$

$\begin{aligned}&=50 \times\left(-2^{2}\right)\\&=50 \times 4\end{aligned}$

= 200 J  

c) To find the work done in this frame by the external force holding the belt :

Frictional force is given by the

$f=\mu_{k} R$

$f=0.2 m g$

$f=0.2 \times 100 \times 10$

$\begin{aligned}&f=0.2 \times 1000\\&f=200 N\end{aligned}$

Retardation $=\frac{f}{m}=\frac{200}{100}=2 \frac{m}{s^{2}}$

The distance that the block moves will be as seen from the frame = s

$\left(u^{\prime}^{2}-v^{\prime 2}\right)=2 a s$

$\left(2^{2}-0^{2}\right)=2 \times 2 s$

$2=2 \times 2 s$

$4 s=4$

$\begin{aligned}&s=\frac{4}{4}\\&s=1 m\end{aligned}$

Work done by the force that accelerates as seen from the frame = fs

$=200 \times 1=200 J$

Work done on the belt to give it a final velocity of 2 m/s $=\frac{1}{2} m v^{\prime 2}$

$=\frac{1}{2} \times 100 \times 2^{2}$

$\begin{aligned}&=50 \times 4\\&=200 J\end{aligned}$

Total work done by the outer force Just like seen from the picture = 200+200 = 400J