A child whose weight is 267 N slides down a 6.5 m playground slide that makes an angle of `20^(@)` with the horizontal. The coefficient of kinetic friction between slide and child is 0.10 (a) How much energy is transferred to thermal energy ? (b) If she starts at the top with a speed of 0.457 m/s, what is her speed at the bottom ?

Answer

To solve the problem, we will break it down into two parts: (a) calculating the energy transferred to thermal energy and (b) finding the final speed of the child at the bottom of the slide.

### Part (a): Energy Transferred to Thermal Energy

1. **Identify the forces acting on the child**:
- Weight of the child, \( W = 267 \, \text{N} \).
- The angle of the slide, \( \theta = 20^\circ \).
- Coefficient of kinetic friction, \( \mu_k = 0.10 \).

2. **Calculate the normal force (N)**:
- The normal force can be calculated using the component of the weight perpendicular to the slide:
\[ N = W \cos(\theta) = 267 \cos(20^\circ) \]

3. **Calculate the frictional force (F_k)**:
- The frictional force is given by:
\[ F_k = \mu_k N = \mu_k (W \cos(\theta)) = 0.10 \times (267 \cos(20^\circ)) \]

4. **Calculate the work done by the frictional force (W_f)**:
- The work done by the frictional force is given by:
\[ W_f = F_k \cdot d = F_k \cdot 6.5 \, \text{m} \]
- Substitute \( F_k \) into this equation:
\[ W_f = 0.10 \times (267 \cos(20^\circ)) \times 6.5 \]

5. **Calculate the numerical value**:
- First, find \( \cos(20^\circ) \approx 0.9397 \):
\[ N \approx 267 \times 0.9397 \approx 250.50 \, \text{N} \]
\[ F_k \approx 0.10 \times 250.50 \approx 25.05 \, \text{N} \]
\[ W_f \approx 25.05 \times 6.5 \approx 162.83 \, \text{J} \]

6. **Final answer for part (a)**:
- The energy transferred to thermal energy is approximately:
\[ \text{Thermal Energy} \approx 162.83 \, \text{J} \]

### Part (b): Final Speed at the Bottom of the Slide

1. **Use the work-energy principle**:
- The work done on the child equals the change in kinetic energy:
\[ W_{\text{net}} = \Delta KE = KE_f - KE_i \]
- Where \( KE_i = \frac{1}{2} m v_i^2 \) and \( KE_f = \frac{1}{2} m v_f^2 \).

2. **Calculate the gravitational force component along the slide**:
- The gravitational force component along the slide is:
\[ F_g = W \sin(\theta) = 267 \sin(20^\circ) \]
- Calculate \( \sin(20^\circ) \approx 0.3420 \):
\[ F_g \approx 267 \times 0.3420 \approx 91.24 \, \text{N} \]

3. **Calculate the net work done (W_net)**:
- The net work done is the work done by gravity minus the work done against friction:
\[ W_{\text{net}} = (F_g \cdot d) - W_f \]
\[ W_{\text{net}} = (91.24 \times 6.5) - 162.83 \]

4. **Calculate the numerical value**:
\[ W_{\text{net}} \approx (91.24 \times 6.5) - 162.83 \approx 593.06 - 162.83 \approx 430.23 \, \text{J} \]

5. **Set up the equation for kinetic energy**:
- The initial kinetic energy:
\[ KE_i = \frac{1}{2} m v_i^2 \]
- The final kinetic energy:
\[ KE_f = \frac{1}{2} m v_f^2 \]
- Since \( m = \frac{W}{g} \approx \frac{267}{9.81} \approx 27.24 \, \text{kg} \), we can substitute \( m \) into the equation:
\[ W_{\text{net}} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]

6. **Solve for \( v_f \)**:
\[ 430.23 = \frac{1}{2} (27.24) v_f^2 - \frac{1}{2} (27.24) (0.457)^2 \]
\[ 430.23 = 13.62 v_f^2 - 2.87 \]
\[ 433.10 = 13.62 v_f^2 \]
\[ v_f^2 \approx \frac{433.10}{13.62} \approx 31.85 \]
\[ v_f \approx \sqrt{31.85} \approx 5.64 \, \text{m/s} \]

### Final Answers:
- (a) The energy transferred to thermal energy is approximately **162.83 J**.
- (b) The speed of the child at the bottom of the slide is approximately **5.64 m/s**.