Q: A ski jumper of mass $$60\ \mathrm{kg}$$ starts at the top of a $$100\ \mathrm{m}$$ long slope inclined at $$\theta = 40^\circ. $$The coefficient of friction is $$\mu = 0.045. $$What is the skier's initial potential energy at the top of the slope? (Assume $$g = 9.8\ \mathrm{m/s^2$}.)$$

$$PE = mgh = 60 \times 100 \times 9.8 = 58,800\ \mathrm{J}$$

Q: As the skier descends the $$100\ \mathrm{m}$$ slope, what is the work done against friction? (Use $$\mu = 0.045$$, $$m = 60\ \mathrm{kg}$$, $$g = 9.8\ \mathrm{m/s^2$}$$, and $$\theta = 40^\circ.)$$

$$W_{fric} = \mu m g \cos\theta \times d = 0.045 \times 60 \times 9.8 \times 0.766 \times 100 \approx 2,030\ \mathrm{J}$$

Q: If the trampoline behaves like a spring, and the first bounce compresses the trampoline by $$\Delta y = 1\ \mathrm{m}$$, what is the spring constant $$k of $$the trampoline? Assume all kinetic energy is stored as elastic potential energy.

$$k = \frac{2KE}{(\Delta y)^2$} = \frac{2 \times 56,770}{1^2$} = 113,540\ \mathrm{N/m}$$

Q: If the skier and the trampoline system lose $$2000\ \mathrm{J}$$ per bounce, what is the total energy lost after 10 bounces?

$$E_{lost} = 2000 \times 10 = 20,000\ \mathrm{J}$$

Q: If the skier's initial velocity at the bottom of the slope is $$v$$, and the skier travels up the icy ramp of $$10\ \mathrm{m} at$$ $$30^\circ$$, what is the change in gravitational potential energy?

$$\Delta PE = m g h = 60 \times 9.8 \times 10 \times \sin 30^\circ = 2,940\ \mathrm{J}$$

Question 1

A ski jumper of mass $$60\ \mathrm{kg}$$ starts at the top of a $$100\ \mathrm{m}$$ long slope inclined at $$	heta = 40^\circ. $$The coefficient of friction is $$\mu = 0.045. $$What is the skier's initial potential energy at the top of the slope? (Assume $$g = 9.8\ \mathrm{m/s^2$}.)$$

Accepted Answer

$$PE = mgh = 60 	imes 100 	imes 9.8 = 58,800\ \mathrm{J}$$

Question 2

As the skier descends the $$100\ \mathrm{m}$$ slope, what is the work done against friction? (Use $$\mu = 0.045$$, $$m = 60\ \mathrm{kg}$$, $$g = 9.8\ \mathrm{m/s^2$}$$, and $$	heta = 40^\circ.)$$

Accepted Answer

$$W_{fric} = \mu m g \cos	heta 	imes d = 0.045 	imes 60 	imes 9.8 	imes 0.766 	imes 100 \approx 2,030\ \mathrm{J}$$

Question 3

After descending the slope, the skier reaches an icy ramp angled at $$30^\circ$$ and $$10\ \mathrm{m}$$ long. If the skier's speed at the bottom of the slope is $$v$$, what is the skier's speed at the top of the ramp, neglecting friction on the ramp?

Accepted Answer

$$v_{top} = \sqrt{v^2$ - 2g h_{ramp}}$$, where $$h_{ramp} = 10 \sin 30^\circ = 5\ \mathrm{m}$$

Question 4

If the skier starts with $$58,800\ \mathrm{J} of $$potential energy and loses $$2,030\ \mathrm{J} to $$friction, what is the skier's kinetic energy at the bottom of the slope?

Accepted Answer

$$KE = 58,800 - 2,030 = 56,770\ \mathrm{J}$$

Question 5

Suppose the skier's velocity at the bottom of the slope is $$v. $$What is the expression for $$v in $$terms of the skier's mass $$m$$, gravitational acceleration $$g$$, height $$h$$, and work done against friction $$W_{fric}$$?

Accepted Answer

$$v = \sqrt{\frac{2(mgh - W_{fric})}{m}}$$

Question 6

If a trampoline is placed at the end of the ramp and the skier lands on it, losing $$2000\ \mathrm{J} of $$energy per bounce, how many times does the skier bounce before all kinetic energy is lost? Assume initial kinetic energy is $$56,770\ \mathrm{J}.$$

Accepted Answer

$$n = \frac{56,770}{2,000} \approx 28$$ bounces

Question 7

If the trampoline behaves like a spring, and the first bounce compresses the trampoline by $$\Delta y = 1\ \mathrm{m}$$, what is the spring constant $$k of $$the trampoline? Assume all kinetic energy is stored as elastic potential energy.

Accepted Answer

$$k = \frac{2KE}{(\Delta y)^2$} = \frac{2 	imes 56,770}{1^2$} = 113,540\ \mathrm{N/m}$$

Question 8

If the skier and the trampoline system lose $$2000\ \mathrm{J}$$ per bounce, what is the total energy lost after 10 bounces?

Accepted Answer

$$E_{lost} = 2000 	imes 10 = 20,000\ \mathrm{J}$$

Question 9

If the skier starts with $$56,770\ \mathrm{J} of $$kinetic energy and loses $$20,000\ \mathrm{J}$$ after 10 bounces, what is the skier's remaining energy?

Accepted Answer

$$E_{rem} = 56,770 - 20,000 = 36,770\ \mathrm{J}$$

Question 10

If the skier's initial velocity at the bottom of the slope is $$v$$, and the skier travels up the icy ramp of $$10\ \mathrm{m} at$$ $$30^\circ$$, what is the change in gravitational potential energy?

Accepted Answer

$$\Delta PE = m g h = 60 	imes 9.8 	imes 10 	imes \sin 30^\circ = 2,940\ \mathrm{J}$$

Question 11

If the skier's velocity at the bottom of the slope is $$v = 43.5\ \mathrm{m/s}$$, what is the skier's kinetic energy?

Accepted Answer

$$KE = \frac{1}{2} m v^2$ = \frac{1}{2} 	imes 60 	imes (43.5)^2$ = 56,677.5\ \mathrm{J}$$

Question 12

If the skier must land at the same altitude ($$y-$$position) as the beginning of the ramp, what should the skier's velocity be at the start of the ramp to reach the end?

Accepted Answer

$$v = \sqrt{2 g h}$$, where $$h = 10 \sin 30^\circ = 5\ \mathrm{m}$$

Question 13

If the skier loses $$2,030\ \mathrm{J} to $$friction on the slope, what percentage of the initial potential energy is lost to friction?

Accepted Answer

$$\frac{2,030}{58,800} 	imes 100\% \approx 3.45\%$$

Question 14

If the skier's mass is doubled, how does the work done against friction change, assuming all other parameters remain the same?

Accepted Answer

The work done against friction doubles, since $$W_{fric} \propto m$$

Question 15

If the coefficient of friction is increased to $$\mu = 0.1$$, what happens to the skier's final speed at the bottom of the slope?

Accepted Answer

The final speed decreases because more energy is lost to friction

Question 16

If the skier starts at rest and slides down the slope, which force does the most work on the skier?

Accepted Answer

Gravity does the most work, converting potential energy to kinetic energy

Question 17

If the skier's initial potential energy is $$58,800\ \mathrm{J}$$ and the total work done by non-conservative forces is $$-2,030\ \mathrm{J}$$, what is the skier's mechanical energy at the bottom of the slope?

Accepted Answer

$$E_{mech} = 58,800 - 2,030 = 56,770\ \mathrm{J}$$

Physics Study Notes: Energy and Motion in Ski Jumping

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