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Chapter 9: Q.56 (page 230)

When a 65kg cheerleader stands on a vertical spring, the spring compresses by 5.5cm. When a second cheerleader stands on the shoulders of the first, the spring compresses an additional 4.5cm. What is the mass of the second cheerleader?

Short Answer

Expert verified

The mass of second cheerleader is53kg.

Step by step solution

01

Content Introduction

Hooke's law says that the force applied on spring is directly proportional to deformation of spring.

FspαxFsp=kx

Here, Fspis force applied on spring, kis spring constant, xis deformation of spring.

02

Content Explanation

Let mass of first cheerleader is m1and mass of second cheerleader is m2.

The value of spring constant when force applied by first cheerleader during compression is 5.5cm

m1g=kx(65kg)(9.81m/s2)=k(5.5cm)(1m100cm)k=116×102N/m

Total force applied on spring is

(m1+m2)g=kx(65kg+m2)(9.81m/s2)=(116×102N/m)(10.0cm)(1m100cm)(65kg+m2)=118kgm2=53kg

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Most popular questions from this chapter

The gravitational attraction between two objects with masses m1 and m2, separated by distance x, is F=Gm1m2/x2 , where G is the gravitational constant.

a. How much work is done by gravity when the separation changes from x1 to x2? Assume x2<x1.

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