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Chapter 6: Problem 51

How much mechanical energy is lost to friction if a 55.0-kg skier slides down a ski slope at constant speed of \(14.4 \mathrm{~m} / \mathrm{s}\) ? The slope is \(123.5 \mathrm{~m}\) long and makes an angle of \(14.7^{\circ}\) with respect to the horizontal.

Short Answer

Expert verified
Answer: The mechanical energy lost to friction is approximately 41483.09 J.

Step by step solution

01

Calculate the vertical displacement of the skier.

To calculate the vertical displacement of the skier, we need to use the length of the slope (\(123.5 \mathrm{~m}\)) and the angle it makes with the horizontal (\(14.7^{\circ}\)): \(h = 123.5 \cdot sin(14.7^{\circ})\) Using the sine function on our calculator, we get: \(h \approx 31.48 \mathrm{~m}\)
02

Determine the gravitational potential energy.

Now we can find the gravitational potential energy (GPE) when the skier is at the top of the slope. The GPE is given by the formula: \(GPE = mgh\) By plugging in the known values, we get: \(GPE = 55.0 \cdot 9.81 \cdot 31.48 \approx 17013.26 \mathrm{~J}\)
03

Calculate the kinetic energy of the skier.

Since the skier slides down the slope at a constant speed (\(14.4 \mathrm{~m/s}\)), we can find their kinetic energy (KE) using the following formula: \(KE = \frac{1}{2}mv^{2}\) By plugging in the known values, we get: \(KE = \frac{1}{2} \cdot 55.0 \cdot (14.4)^{2} \approx 6471.04 \mathrm{~J}\)
04

Calculate the total initial mechanical energy.

Now we can sum up the gravitational potential energy and the kinetic energy to find the initial total mechanical energy: \(Total\_ME = GPE + KE\) \(Total\_ME \approx 17013.26 + 6471.04 \approx 23484.30 \mathrm{~J}\)
05

Determine the work done by friction.

To find the work done by friction, we need to first find the normal force and then use the work-energy theorem. The normal force (\(F_N\)) can be found as follows: \(F_N = mg \cdot cos(14.7^{\circ})\) By plugging in the known values, we get: \(F_N \approx 55.0 \cdot 9.81 \cdot cos(14.7^{\circ}) \approx 526.74 \mathrm{~N}\) Now we can use the work-energy theorem to find the work done by friction: \(W_{friction} = F_{friction} \cdot d \cdot cos(180^{\circ})\) Since the friction force is equal to the normal force, we have: \(W_{friction} = 526.74 \cdot 123.5 \cdot (-1) \approx -64967.39 \mathrm{~J}\)
06

Calculate the mechanical energy lost to friction.

To find the mechanical energy lost to friction, we subtract the work done by friction from the initial total mechanical energy: \(ME_{lost} = Total\_ME - |W_{friction}| \) \(ME_{lost} \approx 23484.30 - 64967.39 \approx -41483.09 \mathrm{~J}\) The negative value indicates that the mechanical energy is indeed being lost to friction. Therefore, the mechanical energy lost to friction is approximately 41483.09 J.

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Most popular questions from this chapter

A spring with a spring constant of \(500 . \mathrm{N} / \mathrm{m}\) is used to propel a 0.500 -kg mass up an inclined plane. The spring is compressed \(30.0 \mathrm{~cm}\) from its equilibrium position and launches the mass from rest across a horizontal surface and onto the plane. The plane has a length of \(4.00 \mathrm{~m}\) and is inclined at \(30.0^{\circ} .\) Both the plane and the horizontal surface have a coefficient of kinetic friction with the mass of \(0.350 .\) When the spring is compressed, the mass is \(1.50 \mathrm{~m}\) from the bottom of the plane. a) What is the speed of the mass as it reaches the bottom of the plane? b) What is the speed of the mass as it reaches the top of the plane? c) What is the total work done by friction from the beginning to the end of the mass's motion?

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