Chapter 6, Problem 11Q

A hill has a height h. A child on a sled (total mass m) slides down starting from rest at the top. Does the speed at the bottom depend on the angle of the hill if (a) it is icy and there is no friction, and (b) there is friction (deep snow)? Explain your answers.

To determine

Whether the speed at the bottom depends on the angle of the hill if it’s icy and there is no friction.

Answer to Problem 11Q

Solution:The speed $\left(v_f\right)$ at the bottom of the hill does not depend on the angle of the hill if there is no friction.

Explanation of Solution

Given:

Hill has a height $\left(h\right)$ .The child on a sled of total mass $\left(m\right)$ slides down starting from rest $\left(v_o\right)$ at the top. There is no friction.

Formula used: If there is no friction, conservation of mechanical energy $\left(E\right)$ can be written as:

  $E=\Delta K+\Delta U=0$  (1)

Where:

• $\Delta K$ is a change of kinetic energy and is defined as:

  $\Delta K=K_f-K_o=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_o^2$  (2)

• $\Delta U$ is a changein potential energy. In this case, we have gravitational potential energy $U_g$, therefore:

  $\Delta U_g=U_f-U_o=mg{h}_f-mg{h}_o$  (3)

Calculation:

According to equations (2) and (3), equation (1) can be written as:

  $K_f+U_f=K_o+U_o$  (4)

or

  $\frac{1}{2}m{v}_f^2+mg{h}_f=\frac{1}{2}m{v}_o^2+mg{h}_o$  (5)

According to given data: $h_f=0,v_o=0$ and $h_o=h$

Then, equation (5) can be written as:

  $\frac{1}{2}m{v}_f^2=mgh$  (6)

Isolating $v_f$ from equation (6) we obtain:

  $v_f=\sqrt{2gh}$  (7)

Conclusion:

According to equation (7), the speed $\left(v_f\right)$ at the bottom of the hill does not depend on the angle of the hill if there is no friction.

To determine

Whether the speed at the bottom depend on the angle of the hill if there is friction (deep snow)?

Answer to Problem 11Q

Solution:The speed $\left(v_f\right)$ at the bottom of the hill depends on the angle $\left(\theta \right)$ of the hill if there is friction.

Explanation of Solution

Given:

Hill has a height $\left(h\right)$ .The child on a sled of total mass $\left(m\right)$ slides down starting from rest $\left(v_o\right)$ at the top. There is friction $\left({\mu }_k\right)$ .

Formula used: If there is friction, conservation of mechanical energy $\left(E\right)$ can be written as:

  $\frac{1}{2}m{v}_f^2+mg{h}_f=\frac{1}{2}m{v}_o^2+mg{h}_o+F_f.d$  (8)

Where:

• $F_f$ is the friction force
• $d$ is the distance along the path traveled by the object

Calculation:

According to data given info: $h_f=0,v_o=0$ and $h_o=h$ then, equation (8) can be written as:

  $\frac{1}{2}m{v}_f^2=mgh+F_{f}d$  (9)

The distance traveled $d$ by the child is:

  $d=h\sin (\theta )$  (10)

And, the friction force $F_f$ is:

  $F_f={\mu }_{k}mg\cos (\theta )$  (11)

Inserting (10) and (11) into equation (9):

  $\frac{1}{2}m{v}_f^2=mgh+{\mu }_{k}mg\cos (\theta )\sin (\theta )$  (12)

Isolating $v_f$ from equation (12):

  $v_f=\sqrt{gh\left[2+{\mu }_k\sin (2\theta )\right]}$  (13)

Conclusion:

According to equation (13), the speed $\left(v_f\right)$ at the bottom of the hill depends on the angle $\left(\theta \right)$ of the hill if there is friction.