Question

To stretch a spring 8.00cm from its unstretched length, 16.0J of work must be done.A)What is the force constant of this spring?B)What magnitude force is needed to stretch the spring 8.00cm from its unstretched length?c) How much work must be done to compress this spring 4.00cm from its unstretched length?D) What force is needed to stretch it this distance?

Answer 1

A) 5000 N/m

The force constant of the spring can be found by using the expression for the elastic potential energy stored in the spring (which is equal to the work done on it):

$W=U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching/compression of the spring

In this problem, we have

W = 16.0 J is the work done

x = 8.00 cm = 0.08 m is the stretching

Substituting into the formula and re-arranging it, we find

$k=\frac{2W}{x^2}=\frac{2(16.0 J)}{(0.08 m)^2}=5000 N/m$

B) 400 N

The magnitude of the force needed to stretch the spring by x = 8.00 cm = 0.08 m is given by Hook's law:

$F=kx$

where k=5000 N/m as we found previously. Substituting x=0.08 m, we find:

$F=(5000 N/m)(0.08 m)=400 N$

C) 4 J

The work done to compress the spring by x=4.00 cm=0.04 m is given by the same formula used for part A:

$W=\frac{1}{2}kx^2$

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

$W=\frac{1}{2}(5000 N/m)(0.04 m)^2=4 J$

D) 200 N

As we did in part B), the force needed to stretch this distance is given by Hook's law:

$F=kx$

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

$F=(5000 N/m)(0.04 m)=200 N$