Question

A child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a speed of 1.10 m/s. The frictional force acting on the sled is one‑fifth of the combined weight of the child and the sled. If she travels for a distance of 22.5 m and her speed at the bottom is 3.70 m/s, calculate the angle that the slope makes with the horizontal.

Answer 1

Answer:

θ = 13.16 °

Explanation:

Lets take mass of child = m

Initial velocity ,u= 1.1 m/s

Final velocity ,v=3.7 m/s

d= 22.5 m

The force due to gravity along the incline plane = m g sinθ

The friction force = (m g)/5

Now from work power energy

We know that

work done by all forces = change in kinetic energy

( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²

(2  g sinθ - ( 2 g)/5 ) d = v² -  u²

take g = 10 m/s²

(20 sinθ - ( 20)/5 ) 22.5 = 3.7² -  1.1²

20 sinθ - 4 =12.48/22.5

θ = 13.16 °