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Chapter 6: Q103P (page 148)

A certain string can withstand a maximum tension of 40 Nwithout breaking. A child ties a 0.37 kgstone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 M, slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? (b) What is the speed of the stone as the string breaks?

Short Answer

Expert verified

a)40Nb)9.5m/s

Step by step solution

01

Given

Max.TensionforceT=40NMassofthestoneM=0.37kgRadiusofthecircleR=0.91m

02

Understanding the concept

This problem is based on the Newton’s second law of motion. Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.

Formula:

F=ma

03

Calculate the stone on its path when the string breaks

(a)

The tension is greatest at the bottom of the swing.

By Newton’s second law:

T-mg=mv2RT=mg+v2R

Increasing the speed eventually leads to the tension at the bottom of the circle reaching that breaking value of 40 N.

04

Calculate the speed of the stone as the string breaks

(b)

Solving the above equation for the speed, we find

v=RTm-g=0.91m40N0.37kg-9.8m/s2=9.5m/s

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Most popular questions from this chapter

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches30°, the box starts to slip, and it then slides 2.5 mdown the plank in 4.0 sat constant acceleration. What are

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