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Chapter 9: Q.59 (page 230)

A horizontal spring with spring constant 250N/m is compressed by 12cm and then used to launch a 250g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box鈥檚 launch speed?

Short Answer

Expert verified

The launch speed of the box is3.7m/s.

Step by step solution

01

Content Introduction

The work energy theorem is

W=鈭�KFnet.d=鈭�K

Considering the following diagram,

The force of kinetic friction is

fK=渭KnfK=渭Kmg

The net force on the mass is calculated as:

Fnet=fsp-fk

Therefore, the total work done on the box during the launch is :

W=鈭�00.12FnetdxW=鈭�00.12(Fsp-fK)dxW=12kx2-渭KmgxW=12(250N/m)(0.12m)2-(0.23)(250g脳1kg1000g)(9.8m/s2)(0.12m)W=1.73J

02

Content Explanation

Now apply the work energy theorem,

W=鈭�KW=12m(vf2-vi2)1.73J=12mvf2vf=2脳1.73J250g脳1kg1000gvf=3.7m/s

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