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

Here,

m = 100 kg

u = 2.0 m/s

v = 0

μk = 0.2

 

a) Internal energy of the belt-block system will decrease when the block will lose its KE in heat due to friction. Thus,

KE lost = `1/2m u^2-1/2mnu^2`

`1/2m(u^2-nu^2)`

`1/2xx100xx(2^2-0^2)`

= 200 J

 

b) Velocity of the frame is given by

u_f = 2.0 m/s

u’ = u - u_f = 2 – 2= 0

v’ = 0 – 2 = -2 m/s

KE lost = `1/2m u'^2-1/2mnu'^2`

`=1/2m(0^2-nu'^2)`

`=1/2xx100xx(0^2-2^2)`

= 200 J

 

c) Force of friction is given by

`f=mu_k^""R`

`rArrf=0.2xxmg=0.2xx100xx10=200N

`"Retardation"=f/m=200/100=2ms^-2`

Distance moved by the block will be as seen from the frame = s

`nu'^2-u'^2=2as`

`rArr2^2-0^2=2xx2s`

`rArrs=1"m"`

Work done by the force responsible for accelaration as seen from the frame = fs

= 200 x 1 = 200J

Work done by the belt to give it a final velocity of 2 m/s

`=1/2mnu'^2`

`=1/2xx100xx(2)^2`

= 200J

Total work done by external force as seen from the frame 200 + 200 = 400J