Question

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 450 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 30 centimeters to 60 centimeters?

Answer 1

Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

$F =450N$

$x_1 = 30cm$

$x_2 = 60cm$

Required

The work done

First, we calculate the spring constant (k)

$F = kx_1$

$450N = k *30cm$

$k = \frac{450N}{30cm}$

$k =15N/cm$

So:

$F = kx_1$

$F(x) = 15x$

The work done using Hooke's law is:

$W =\int\limits^a_b {F(x)} \, dx$

This gives:

$W =\int\limits^{60}_{30} {15x} \, dx$

Rewrite as:

$W =15\int\limits^{60}_{30} {x} \, dx$

Integrate

$W =15 \frac{x^2}{2}|\limits^{60}_{30}$

This gives:

$W =15 *\frac{60^2 - 30^2}{2}$

$W =15 *\frac{2700}{2}$

$W =15 *1350$

$W =20250N-cm$

Convert to Nm

$W =\frac{20250Nm}{100}$

$W =202.50Nm$