Questions: Child goes down an 8 meter slide at an angle of 30°. The coefficient of kinetic friction is 0.35. If the child starts from rest, what is his speed when he reaches the bottom of the slide?

Child goes down an 8 meter slide at an angle of 30°. The coefficient of kinetic friction is 0.35. If the child starts from rest, what is his speed when he reaches the bottom of the slide?

Transcript text: Child goes down an 8 meter slide at an angle of $30^{\circ}$. The coefficient of kinetic friction is 0.35 . If the child starts from rest, what is his speed when he reaches the bottom of the slide?

Solution

Solution Steps

Step 1: Identify the forces acting on the child

The forces acting on the child are gravity, normal force, and friction. The gravitational force can be decomposed into two components: one parallel to the slide and one perpendicular to the slide.

Step 2: Calculate the gravitational force components

The gravitational force $ F_g $ is given by $ F_g = mg $, where $ m $ is the mass of the child and $ g $ is the acceleration due to gravity (9.81 m/s$^2$).

The component of the gravitational force parallel to the slide is: \[ F_{\parallel} = mg \sin(30^\circ) \]

The component of the gravitational force perpendicular to the slide is: \[ F_{\perp} = mg \cos(30^\circ) \]

Step 3: Calculate the normal force

The normal force $ F_N $ is equal to the perpendicular component of the gravitational force: \[ F_N = mg \cos(30^\circ) \]

Step 4: Calculate the frictional force

The frictional force $ F_f $ is given by: \[ F_f = \mu_k F_N = \mu_k mg \cos(30^\circ) \] where $ \mu_k $ is the coefficient of kinetic friction (0.35).

Step 5: Calculate the net force along the slide

The net force $ F_{\text{net}} $ acting along the slide is the difference between the parallel component of the gravitational force and the frictional force: \[ F_{\text{net}} = F_{\parallel} - F_f \] \[ F_{\text{net}} = mg \sin(30^\circ) - \mu_k mg \cos(30^\circ) \]

Step 6: Calculate the acceleration

Using Newton's second law $ F = ma $, the acceleration $ a $ of the child along the slide is: \[ a = \frac{F_{\text{net}}}{m} \] \[ a = g \sin(30^\circ) - \mu_k g \cos(30^\circ) \]

Step 7: Calculate the final speed

The child starts from rest, so the initial velocity $ u = 0 $. Using the kinematic equation $ v^2 = u^2 + 2as $, where $ s $ is the length of the slide (8 meters), we can solve for the final speed $ v $: \[ v^2 = 0 + 2as \] \[ v = \sqrt{2as} \]

Substitute the values: \[ a = 9.81 \sin(30^\circ) - 0.35 \times 9.81 \cos(30^\circ) \] \[ a = 9.81 \times 0.5 - 0.35 \times 9.81 \times \frac{\sqrt{3}}{2} \] \[ a = 4.905 - 0.35 \times 9.81 \times 0.8660 \] \[ a = 4.905 - 2.973 \] \[ a = 1.932 \, \text{m/s}^2 \]

Now, calculate the final speed: \[ v = \sqrt{2 \times 1.932 \times 8} \] \[ v = \sqrt{30.912} \] \[ v = 5.560 \, \text{m/s} \]