Question

A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The coefficient of friction between the rollerblader and the floor is 0.25 when she applies the brakes. Ignore the friction between the rollerblader and the inclined surface.
Use Newton's second law to estimate the distance she will move before stopping.

Answer 1

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

$\dfrac{1}{2}mu^2 = m g d sin \theta$

$u^2 = 2 g d sin30^0$

$u= \sqrt{2\times 9.8 \times 10 sin30^0}$

u = 9.89 m/s

Using second law of motion  

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,  

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m