4. a) When a car drives at constant speed in a circle on a flat road, the work done by the staticfriction between the tires and the road is zero. Explain why. (it has something to do withdirections)c) When using the formula K, + U, + Wnc = K, + U2, we do not include the work done bygravity or a spring in the term Wnc. Why not? Doesn't that leave out the effect of those forces? Iam looking for two parts to the answer.d) In a collision between two objects where the objects only interact with each other, ismomentum sometimes or always conserved, and is kinetic energy sometimes or alwaysconserved? If it is sometimes, be clear on when it is conserved.

a) The car is moving at a constant speed in a circle. The centripetal force required to keep the car…

Answered: 4. a) When a car drives at constant…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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You slide down a snowy hill on a toboggan from a height of 7 m. The hill has a constant slope such that your speed remains constant at 5 m/s throughout the entire journey. Air resistance is negligible. Which of the following statements is true? a. The magnitude of the work done by gravity is larger than the magnitude of the work done by friction. b. The work done by friction is equal to the change in the gravitational potential energy. c. The work done by friction cannot be determined. d. Friction is not significant because you actually move. e. None of the above are true.
Let's say the Earth moves around the Sun in a circular orbit with a constant speed (neither is completely true, but they're very close, so let's just say it's true). Select the most correct statement about the work done on the Earth. A. The work done on the Earth is negative, since it is technically "falling" into the Sun, so its energy is decreasing. B. The work done on the Earth is zero, since the force of gravity is always perpendicular to the Earth'sdirection of motion.C. The work done on the Earth is zero, since the centripetal force cancels out the Sun's force of gravity.D. The work done on the Earth is positive, since it is technically "falling" into the Sun, its potential energy is decreasing, meaning its kinetic energy is increasing.E. The work done on the Earth is positive, since there is a net force on the Earth (the Sun's force of gravity) which is constantly changing the direction of the Earth's velocity.
3. A 10.0-kg box is pulled by a horizontal wire in a circle on a rough horizontal surface for which the coefficient of kinetic friction is 0.250. Calculate the work done by friction during one complete circular trip if the radius is: a. 2.00 m and b. 4.00 m. c. On the basis of the results you just obtained, would you say that friction is a conservative or nonconservative force? Explain.
A Ferris wheel is a vertical, circular amusement ride with radius, R = 5 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates so that riders move at a constant speed, v = 5 m/s. Consider a rider whose mass, m = 70 kg. Assume that the acceleration due to gravity, g = 10 m/s₂. H B * L. How much work does the force exerted by the seat do on the rider as they move from point E to point ? O None of these, because the rider's speed does not change O 3500) O-3501 O-3500
The last word of the sentence in the picture is "particle" a. Find the work done by the force using a dot product. b. Find the angle between the force and displacement.
In the attached image, the mass is 3.20 kg, and theta=37 degrees. The surface does have friction, with static coefficient=0.45 and kinetic coefficient=0.39.If a person takes hold of the rope and, pulling at this angle with a force of T=23 N, drags the box 3.8 meters... a. How much work is done on the box by person pulling? b. How much work is done on the box by the force of friction? c. What is the net work done on the box?
n )i – Î - (183 N )ĵor -> A floating ice block is pushed through a displacement d = (27.0 mî - (24.0 m )ĵ along a 24.0 m )j along a straight embankment by rushing water, which exerts a force F = (298 N )î – (183 N )ĵ on the block. How much work does the force do on the block during the displacement? Number i Units
a) A person is riding on a Ferris wheel of radius R. He starts at the lowest point of the wheel. When the wheel makes one complete revolution, is the net work done by the gravitational force positive, zero or negative? Do you need to know how the speed of the person changed before you can answer the question? b) Suppose the person started at the highest point of the wheel. What is the net work done by the gravitational force when he reaches the lowest point?
Consider a satellite in a circular orbit above Earth’s surface. In Chapter we will learn that the force of gravity changes only the direction of motion of a satellite in circular motion (and keeps it in a circle); it does NOT change the satellite’s speed. Work done on the satellite by the gravitational force is zero. What is your explanation?
When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of the air. Model a cyclist as having cross-section area 0.42 m2 and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90. a.) What is the cyclist's power output while riding at a steady 7.3 m/s (16 mph)? Use 1.3 kg/m3 as a value for the density of air. P= b.) Metabolic power is the rate at which your body "burns" fuel to power your activities. For many activities, your body is roughly 25% efficient at converting the chemical energy of food into mechanical energy. What is the cyclist's metabolic power while cycling at 7.3 m/s? PM= c.) The food calorie is equivalent to 4190 J. How many calories does the cyclist burn if he rides over level ground at 7.3 m/s for 2.0 h? c=
C 4. A box is being pulled across a rough floor with a force P. What can you say about the work done by pulling force, normal force, friction, and gravitational force? 0, W by normal force > 0, Wby friction 0, Why normal force 0, W by normal force= D. Wby pulling force <0, Why normal force = 0, W by friction <0, W by gravitational force = 0 0, W by friction <0. Wby gravitational force = 0
While leaning out a window that is 6.4 m above the ground, you drop a 0.60 kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.9 m above the ground. Determine the following. a)the amount of work done (in J) by the force of gravity on the ball. b) the gravitational potential energy (in J) of the ball-earth system, relative to the ground when it is released. c) the gravitational potential energy (in J) of the ball-earth system, relative to the ground when it is caught. d) the ratio of the change (PEf − PE0) in the gravitational potential energy of the ball-earth system to the work done on the ball by the force of gravity

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