Question

a suitcase is gently dropped on a conveyor belt moving with a speed of 4.9m/s.if the coefficient of friction between the suitcase and the belt is 0.7, then how far will the suitcase move on the belt before coming to rest?​

Answer 1

Given:

a suitcase is gently dropped on a conveyor belt moving with a speed of 4.9m/s.if the coefficient of friction between the suitcase and the belt is 0.7.

To find:

Distance after which the belt will come to rest.

Calculation:

The whole mechanics can be imagined in the reference frame of the belt. When the suitcase has been released on the belt , initially it is stationary. But friction would try to prevent any relative displacement and as a result will make the bag move at 3 m/s such that it comes to rest w.r.t belt

So , considering reference frame to be that of belt.

$\sf{ \therefore \: {v}^{2} = {u}^{2} + 2as}$

$\sf{ = > \: {v}^{2} = {u}^{2} + 2( - \mu g)s}$

$\sf{ = > \: {(0)}^{2} = {(4.9)}^{2} + 2( - \mu g)s}$

$\sf{ = > \: 2 \mu gs = 4.9 \times 4.9}$

$\sf{ = > \: 2 \times (0.7 \times 9.8)s = 4.9 \times 4.9}$

$\sf{ = > \: 4 \times 0.7 \times s = 4.9 }$

$\sf{ = > \: s = \dfrac{4.9}{4 \times 0.7} }$

$\sf{ = > \: s = \dfrac{0.7}{4} }$

$\sf{ = > \: s = 0.175 \: m}$

$\sf{ = > \: s = 17.5 \: cm}$

So, the suitcase will come to rest w.r.t belt after 17.5 cm.

Answer 2

Answer:

s = 1.715 m

Explanation:

Since the suitcase is slipping, deceleration because of friction would be given by:

a=−Mμg / M

a = −μg

a = - 0.7 x 10 = - 7

Using equation of motions we have

v2−u2=2as

In our case,

v = 0 and u = 4.9

Putting the values;

- (4.9)^2 = - 2 x 7x s

s = 1.715 m