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Chapter 6: Problem 63

A car of mass \(987 \mathrm{~kg}\) is traveling on a horizontal segment of a freeway with a speed of \(64.5 \mathrm{mph}\). Suddenly, the driver has to hit the brakes hard to try to avoid an accident up ahead. The car does not have an ABS (antilock braking system), and the wheels lock, causing the car to slide some distance before it is brought to a stop by the friction force between the car's tires and the road surface. The coefficient of kinetic friction is \(0.301 .\) How much mechanical energy is lost to heat in this process?

Short Answer

Expert verified
Answer: The mechanical energy lost to heat in this process is 786909.16 Joules.

Step by step solution

01

Convert the initial velocity to m/s

Given that the initial velocity of the car is 64.5 mph, we need to convert this to meters per second (m/s). Formula to convert mph to m/s is: 1 mph = 0.44704 m/s So, 64.5 mph = 64.5 * 0.44704 m/s = 28.826 m/s
02

Compute initial kinetic energy of the car

We know that the mass of the car (m) is 987 kg and the initial velocity (v) is 28.826 m/s. The initial kinetic energy (KE) of the car can be calculated using the formula: KE = (1/2)mv^2 KE = (1/2) * 987 * (28.826)^2 = 393454.58 Joules
03

Compute the acceleration due to friction

We know that the coefficient of kinetic friction (μ) is 0.301. Since the car is on a horizontal surface, the only force acting on the car is the friction force. We can calculate the acceleration (a) due to friction using the formula: a = μg a = 0.301 * 9.81 = 2.9501 m/s^2
04

Compute the stopping distance

The car comes to a stop, hence the final velocity (vf) is 0 m/s. We can use the equation of motion \(v_f^2 = v_i^2 + 2as\) to find the stopping distance (s). Rearranging the equation for s, we get: s = (vf^2 - vi^2) / (2a) s = (0 - (28.826)^2) / (2 * -2.9501) = 140.98 meters
05

Compute work done by the friction force (W)

Now we will calculate the work done by the friction force (W) using the formula: W = -fs Since the friction force (f) is equal to m * a, W = -(m * a) * s W = -(987 * 2.9501) * 140.98 = -393454.58 Joules
06

Compute the mechanical energy lost to heat

The mechanical energy lost to heat is equal to the difference between the initial kinetic energy of the car and the work done by the friction force. Since the work done by the friction force is negative, the energy lost (ΔE) can be found by: ΔE = KE - W ΔE = 393454.58 - (-393454.58) = 786909.16 Joules The mechanical energy lost to heat in this process is 786909.16 Joules.

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Most popular questions from this chapter

You have decided to move a refrigerator (mass \(=81.3 \mathrm{~kg}\), including all the contents) to the other side of the room. You slide it across the floor on a straight path of length \(6.35 \mathrm{~m}\), and the coefficient of kinetic friction between floor and fridge is \(0.437 .\) Happy about your accomplishment, you leave the apartment. Your roommate comes home, wonders why the fridge is on the other side of the room, picks it up (you have a strong roommate!), carries it back to where it was originally, and puts it down. How much net mechanical work have the two of you done together?

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?

A father exerts a \(2.40 \cdot 10^{2} \mathrm{~N}\) force to pull a sled with his daughter on it (combined mass of \(85.0 \mathrm{~kg}\) ) across a horizontal surface. The rope with which he pulls the sled makes an angle of \(20.0^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.200,\) and the sled moves a distance of \(8.00 \mathrm{~m}\). Find a) the work done by the father, b) the work done by the friction force, and c) the total work done by all the forces.

An arrow is placed on a bow, the bowstring is pulled back, and the arrow is shot straight up into the air; the arrow then comes back down and sticks into the ground. Describe all of the changes in work and energy that occur.

A 0.100 -kg ball is dropped from a height of \(1.00 \mathrm{~m}\) and lands on a light (approximately massless) cup mounted on top of a light, vertical spring initially at its equilibrium position. The maximum compression of the spring is to be \(10.0 \mathrm{~cm}\). a) What is the required spring constant of the spring? b) Suppose you ignore the change in the gravitational energy of the ball during the 10 -cm compression. What is the percentage difference between the calculated spring constant for this case and the answer obtained in part (a)?
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