Stopping Of Block Due To Friction

On the Horizontal Road

A block of mass m is moving initially with velocity u on a rough surface and due to friction, it comes to rest after covering a distance S.

• Distance travelled before coming to rest (S):

F=ma=μRma=μmga=μgV2=u2−2asS=u22μg=P22μm2g

Where:

a= acceleration μ= coefficient of friction S= distance travelled g= gravity u= initial velocity V= finally velocity P= initial mometum=mu

• Time taken to come to rest:

From equation, v=u−at⇒0=u−μgt[ As v=0,a=μg]
∴t=uμg

• The force of friction acting on the body:

F=maF=m(v−u)tF=mut[ As v=0]F=μmg[ As t=uμg]

On the Inclined Road

a=g[sin⁡θ+μcos⁡θ]V2=u2−2aS0=u2−2g[sin⁡θ+μcos⁡θ]SS=u22g(sin⁡θ+μcos⁡θ)

Where:

S= distance travelled μ= coefficient of friction V= Final velocity u= Initial velocity

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Solved Examples Based on Stopping of Block Due to Friction

Example 1: A 2 kg mass starts from rest on an inclined smooth surface with an inclination 30∘ and a length 2 m. How much will it travel (in meters) before coming to rest on a surface with a coefficient of friction of 0.25?

1) 4

2)6

3)8

4)2

Solution:

v2=u2+2as=0+2×gsin⁡30×2

Let it travel a distance ' S ' before coming to rest on a horizontal surface
S=v22μg=202×0.25×10=4 m

Hence, the answer is option (1).

Example 2: A body is moving with a speed of 12m/s and the coefficient of friction between the ground and the body is 0.4. The distance travelled by the body (in meters) before coming to rest is.

1) 18

2)10

3)15

4)20

Solution:

f=μmg
due to retardation
μmg=ma⇒a=μg=0.4∗10=4 m/s2
now for V=0
V2=u2−2as⇒0=u2−2asS=u22×a⇒12∗122∗4=18 m

Hence, the answer is option (1).

Example 3: A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms−1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is:
[ use g=9.8 ms−2]

1) 4.9 m
2) 9.8 m
3) 12.5 m
4) 19.6 m

Solution:

f=μN=μmga=μg=4.9 m/s2V2=u2+2as0=(9.8)2+2(−4.9)ss=9.8 m
The distance covered by the block before coming to rest is $\mathrm{ 9.8\, m}$

Hence, the correct option is (2)

Example 4: A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during the slipping motion, is : [Take g=10 m/s−2 ]

1) 2 m
2) 0.5 m
3) 3.2 m
4) 0.8 ms

Solution:

f=μmga=fm=μg

The bag comes to rest when the speed of the bag is the same as the conveyor belt ie., 4 m/s
v2=u2+2 as (4)2=0+2(0.4×10)sS=2 m

The distance travelled by the bag on the belt during slipping motion is 2 m

Hence 1 is the correct option.

Example 5: A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the inclined plane with horizontal is θ. The magnitude of the contact force will be:

1) Mg
2) Mgcos⁡θ
3) Mgsin⁡θ+Mgcos⁡θ
4) Mgsin⁡θ1+μ

Solution:

For block moving at constant velocity

∑Fx=0−mgsin⁡Θ+μMgcos⁡Θ=0tan⁡Θ=μf=Mgsin⁡Θ=μMgcos⁡Θ→(1) Contact force =f2+N2(Mgsin⁡Θ)2+(Mgcos⁡Θ)2 Contact force =Mg

Hence, the answer is option (1).

Summary

When a moving block encounters friction, the force of friction opposes its motion, gradually reducing its speed until it comes to a complete stop. The stopping distance depends on the initial speed of the block and the coefficient of friction between the block and the surface. Understanding this interaction is crucial for applications requiring precise control of motion and stopping.