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

Answer:

A)s = 104.16 m

b)s= 104.16 m

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

$- Fr.s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Negative sign because force act opposite to the displacement.

$- \mu\ m\ g \ s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$-\mu\ g \ s=\dfrac{1}{2}v^2-\dfrac{1}{2}u^2$

$-0.3\times 10\times \ s=\dfrac{1}{2}\times 0^2-\dfrac{1}{2}\times 25^2$

- 3 x 2 x s = - 625

s= 104.16 m