Question

7. A 25 kg child slides down a 6.5 m playground slide that makes an angle of 20o with horizontal. The coefficient of kinetic friction between slide and child is 0.10.

Find the following:

a. What kind of energy is present on the child at the top of the slide? At the bottom of the slide?

b. How much potential energy does the child have?

c. How much kinetic energy does the child have?

d. How much energy is transferred to thermal energy?

e. If she starts at the top with A SPEED OF 0.50 m/s, what is her speed at the bottom?

Answer 1

a) If the child starts from rest, she has only potential energy at the top of the slide. At the bottom she has only kinetic energy, if the zero potential is chosen to be at the level of the bottom of the slide.

b) At the top she has

$$E_p = 25kg\cdot 9.8m/s^2\cdot 6.5m\cdot \sin20\degree \approx 540J$$

c) The amount of the kinetic energy determined by the change of the potential energy and work $W_f$ done by friction force, according to the energy-work theorem:

$$E_k = E_p-W_f$$

The work done by friction force is

$$W_f = F_fd$$

where $F_f = \mu mg\cos \theta$ is the friction force (here $\mu = 0.1,m = 25kg,g = 9.8,\theta = 20\degree$) and $d = 6.5m$. Thus, obtain:

$$E_k = 540-0.1\cdot 25\cdot 9.8\cdot \cos20\degree \approx 517J$$

d) The thermal energy is the difference between the initial potential and final kinetic energy:

$$E_t = E_p-E_k \approx 23J$$

e) According to the energy-work theorem:

$$E_f = E_i + E_p-W_f$$

where $E_i = \dfrac{mv^2}{2}$ and $v = 0.50m/s$. Thus, obtain:

$$E_f = \dfrac{25\cdot 0.5^2}{2} + 517 \approx 520J$$