Question

32. (FR) A mass moving at 25 m/s slides along a rough horizontal surface. The coefficient of friction is 0.30. (A) Use force and kinematics to find stopping distance. (B) Using energy solve for the stopping distance.

Answer 1

A)s = 104.16 m

b)s= 104.16 m

Step-by-step explanation:

Given that

u = 25 m/s

μ = 0.3

The friction force will act opposite to the direction of motion.

Fr= μ m g

Fr= - m a

a=acceleration

μ m g = - m a

a= - μ g

a= - 0.3 x 10 m/s² ( take g= 10 m/s²)

a= - 3 m/s²

The final speed of the mass is zero ,v= 0

We know that

v² = u² +2 a s

s=distance

0² = 25² - 2 x 3 x s

625 = 6 s

s = 104.16 m

By using energy conservation

Work done by all the forces =Change in the kinetic energy

`- Fr.s=(1)/(2)mv^2-(1)/(2)mu^2`

Negative sign because force act opposite to the displacement.

`- \mu\ m\ g \ s=(1)/(2)mv^2-(1)/(2)mu^2`

`-\mu\ g \ s=(1)/(2)v^2-(1)/(2)u^2`

`-0.3* 10* \ s=(1)/(2)* 0^2-(1)/(2)* 25^2`

- 3 x 2 x s = - 625

s= 104.16 m