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Chapter 10: Q.15 (page 256)

How far must you stretch a spring with k = 1000 N/m to store 200 J of energy?

Short Answer

Expert verified

As the all we solve above from that we have , the spring's stretched distance is0.63m.

Step by step solution

01

Introduction

You must exert effort in order to compress or stretch a spring. You must apply a force to the spring that is the same magnitude as the force the spring applies to you, but in the opposite direction. Fext = kx is the force you apply in the direction of the displacement x from its equilibrium position.

02

Explanation

In a compressed or extended spring, the potential energy stored is,

Us=12k(螖s)2

We have, k= spring constant and螖sis the length of the spring compressed or stretched.

Rewriting equation 螖s,

螖s=2Usk

Putting 200Jfor Usand 1000N/mfor k, solve for 螖s.

螖s=2(200J)1000N/m

=0.63m

As the all we solve above from that we have , the spring's stretched distance is 0.63m.

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Most popular questions from this chapter

The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that the leg bones of a healthy, 60 kg human can absorb about 200 J. From what maximum height could a 60 kg person jump and land rigidly upright on both feet without breaking his legs? Assume that all energy is absorbed by the leg bones in a rigid landing

A 10 kg runaway grocery cart runs into a spring with spring constant 250 N/m and compresses it by 60 cm. What was the speed of the cart just before it hit the spring?

a. A 50gice cube can slide without friction up and down a 30oslope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10cm. The spring constant is 25N/m. When the ice cube is released, what total distance will it travel up the slope before reversing direction?

b. The ice cube is replaced by a 50gplastic cube whose coefficient of kinetic friction is 0.20. How far will the plastic cube travel up the slope? Use work and energy.

A system of two objects has 鈭�Ktot=7Jand鈭�Uint=-5J.

a. How much work is done by interaction forces?

b. How much work is done by external forces?

In Problems 66 through 68you are given the equation used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation.

b. Draw the before-and-after pictorial representation.

c. Finish the solution of the problem.

12(1500kg)(5.0m/s)2+(1500kg)9.80m/s2(10m)=12(1500kg)vi2+(1500kg)9.80m/s2(0m)
See all solutions

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