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Chapter 6: Problem 36

A 75.0 -kg skier starts from rest and slides down a 32.0 -m frictionless slope that is inclined at an angle of \(15.0^{\circ}\) with the horizontal. Ignore air resistance. (a) Calculate the work done by gravity on the skier and the work done by the normal force on the skier. (b) If the slope is not frictionless so that the skier has a final velocity of \(10.0 \mathrm{m} / \mathrm{s},\) calculate the work done by gravity, the work done by the normal force, the work done by friction, the force of friction (assuming it is constant), and the coefficient of kinetic friction.

Short Answer

Expert verified
Question: Calculate the work done by gravity, normal force, and friction in both scenarios - a frictionless slope and a non-frictionless slope with a given final velocity of 10 m/s - for a skier with a mass of 70kg sliding down a 32 m long slope at an angle of 15 degrees. Answer: For Scenario (a) - Frictionless slope: 1. Gravitational Potential Energy (MPE) = ________ J 2. Work done by gravity (W_g) = ________ J 3. Work done by the normal force (W_NF) = 0 J For Scenario (b) - Non-frictionless slope with the given final velocity: 4. Work done by gravity (W_g) = ________ J 5. Final Kinetic Energy of the skier (KE) = ________ J 6. Work done by friction (W_f) = ________ J 7. Force of friction (F_f) = ________ N 8. Coefficient of kinetic friction (μ_k) = ________

Step by step solution

01

Calculate the gravitational potential energy (MPE) of the skier at the top of the slope

To find the GPE, use the formula \(MPE = mgh\), where \(m\) is the mass of the skier, \(g\) is the acceleration due to gravity (\(9.81\,\text{m/s}^2\)), and \(h\) is the vertical height of the slope. To find the height (\(h\)), use the trigonometric relationship \(\sin(15^{\circ}) = \frac{h}{32\,\text{m}}\). Solve for \(h\).
02

Calculate the work done by gravity

Since the slope is frictionless, all of the skier's gravitational potential energy is converted into kinetic energy, and the work done by gravity is equal to the change in the skier's kinetic energy (\(W_g = \Delta KE\)). So, \(W_g = MPE\).
03

Calculate the work done by the normal force

Since the normal force acts perpendicular to the direction of motion (along the slope), the work done by the normal force is zero (\(W_{NF} = 0\)). For scenario (b) - Non-frictionless slope with the given final velocity:
04

Calculate the work done by gravity

The work done by gravity can be calculated using the same method as in Step 2, as \(W_g = MPE\).
05

Determine the final kinetic energy of the skier

Use the formula \(KE = \frac{1}{2}mv^2\), with the final velocity given as \(10.0\,\text{m/s}\), to find the final kinetic energy of the skier.
06

Calculate the work done by friction

The work-energy principle states that the net work done on an object equals the change in its kinetic energy. In this case, the net work is due to gravity and friction (\(W = W_g + W_f\)). Since \(W = \Delta KE\) and we have the values for \(W_g\) and \(\Delta KE\), solve for the work done by friction (\(W_f\)).
07

Calculate the force of friction

Since the work done by friction is equal to the force of friction multiplied by the distance the skier slides along the slope, use the formula \(W_f = F_f \cdot d\), where \(d\) is the distance of the slope, to find the force of friction (\(F_f\)).
08

Calculate the coefficient of kinetic friction

The force of friction is related to the normal force by the equation \(F_f = \mu_k F_N\), where \(\mu_k\) is the coefficient of kinetic friction and \(F_N\) is the magnitude of the normal force. The normal force is equal to the component of the gravitational force perpendicular to the slope, which can be calculated as \(F_N = mg\cos(15^{\circ})\). Solve for \(\mu_k\) using the values of \(F_f\) and \(F_N\).

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