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Physics For Scientists & Engineers
Found in: Page 237
Physics For Scientists & Engineers

Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

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Short Answer

A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s. The coefficient of kinetic friction between sled and ice is 0.100. Use energy considerations to find the distance the sled moves before it stops.

The distance the sled moves before it stops is d=2.04 m

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of the non-isolated system

A non-isolated system is one for which energy crosses the boundary of the system. An isolated system is one for which no energy crosses the boundary of the system.

If a friction force of magnitude fk acts over a distance d within a system, the change in internal energy of the system is

ΔEint=fkd..........8.14

Non isolated System (Energy): The most general statement describing the behavior of a non isolated system is the conservation of energy equation

ΔEsystem=∑T..........8.1

Step 2: Calculating the distance

We could solve this problem using Newton’s second law, but we will use the non isolated system energy model, from using the equation (8.1) and (8.14) here written as -fkd=kf-ki, where the kinetic energy change of the sled after the kick results only from the friction between the sled and ice.

The weight and normal force both act at to the motion, and therefore do no work on the sled. The friction force is

role="math" localid="1663680345416" fk=μkn=μkmgd

Since the final kinetic energy is zero, we have

-fkd=ki12mvi2=μkmg

Solving, we get

d=mvi22fk=mvi22μkmg=vi22μkg

Substituting the values, we get

d=2 m/s220.1009.8 m/s2=2.04 m

Thus, the distance is 2.04 m

Most popular questions for Physics Textbooks

A 10.0-kg block is released from rest at point A in Figure P8.63. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2250 N/m, and compresses the spring 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points B and C.

Heedless of danger, a child leaps onto a pile of old mattresses to use them as a trampoline. His motion between two particular points is described by the energy conservation equation

12(46.0 kg)(2.40 m/s)2+(46.0 kg)(9.80 m/s2)(2.80 m+x)=12(1.94×104 N/m)x2

(a) Solve the equation for x. (b) Compose the statement of a problem, including data, for which this equation gives the solution. (c) Add the two values of x obtained in part (a) and divide by 2. (d) What is the significance of the resulting value in part (c)?

Why is the following situation impossible? A softball pitcher has a strange technique: she begins with her hand at rest at the highest point she can reach and then quickly rotates her arm backward so that the ball moves through a half-circle path. She releases the ball when her hand reaches the bottom of the path. The pitcher maintains a component of force on the 0.180-kg ball of constant magnitude 12.0 N in the direction of motion around the complete path. As the ball arrives at the bottom of the path, it leaves her hand with a speed of 25.0 m/s

A roller-coaster car shown in Figure P8.72 is released from rest from a height h and then moves freely with negligible friction. The roller-coaster track includes a circular loop of radius R in a vertical plane. (a) First suppose the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless. Find the required height h of the release point above the bottom of the loop in terms of R. (b) Now assume the release point is at or above the minimum required height. Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the car’s weight. The normal force on each rider follows the same rule. Such a large normal force is dangerous and very uncomfortable for the riders. Roller coasters are therefore not built with circular loops in vertical planes. Figure P6.17 (page 170) shows an actual design.

A 20.0 kg cannonball is fired from a cannon with muzzle speed of 1000 m/s at an angle of 37° with the horizontal. A second ball is fired at an angle of 90°. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let y=0 at the cannon.
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