# A child starts from rest and slides down a snow-covered hill with a slope angle of 47 degree from...

## Question:

A child starts from rest and slides down a snow-covered hill with a slope angle of {eq}47 ^\circ {/eq} from a height of {eq}1.8\ m {/eq} above the bottom of the hill. The speed of the child at the bottom of the hill is {eq}\displaystyle 3.1\ \rm m/s {/eq}. Find the coefficient of kinetic friction between the hill and the child.

## Work-done by friction:

The force that is provided by the surface on the object while having a relative motion with the surface is known as friction force. And the work done by this force is known as the frictional energy.

Given data

The angle of the slope {eq}(\theta) = 47^\circ {/eq}

height of the plane {eq}(H) = 1.8 \ m {/eq}

Speed of the child at the bottom of the hill {eq}v = 3.1 \ m/s {/eq}

Now, the length of the plane would be

{eq}L = \dfrac{H}{\sin\theta} \\ L = \dfrac{1.8}{\sin47^\circ} \\ L = 2.46 \ m {/eq}

Now, the energy at the top of the hill

{eq}E_{1} = mgH \\ {/eq}

where

• m is the mass of the child

The energy at the bottom of the hill

{eq}E_{2} = \dfrac{1}{2}mv^{2} {/eq}

Now, the lost in the energy will be equal to the friction, therefore

{eq}E_{f} = E_{1} - E_{2} \\ f \times L = (mgH) - (0.5mv^{2}) \\ (\mu*mg\cos\theta) \times L = m [ gH - 0.5v^{2} ] \\ \mu \times 9.8 \cos47^\circ \times 2.46 = (9.8 \times 1.8) - (0.5 \times 3.1^{2}) \\ \mu = 0.781 {/eq}

where

• {eq}\mu {/eq} is the coefficient of kinetic friction
• f is the friction force