Answer:
The work done is 202.50Nm
Step-by-step explanation:
Given
[tex]F =450N[/tex]
[tex]x_1 = 30cm[/tex]
[tex]x_2 = 60cm[/tex]
Required
The work done
First, we calculate the spring constant (k)
[tex]F = kx_1[/tex]
[tex]450N = k *30cm[/tex]
[tex]k = \frac{450N}{30cm}[/tex]
[tex]k =15N/cm[/tex]
So:
[tex]F = kx_1[/tex]
[tex]F(x) = 15x[/tex]
The work done using Hooke's law is:
[tex]W =\int\limits^a_b {F(x)} \, dx[/tex]
This gives:
[tex]W =\int\limits^{60}_{30} {15x} \, dx[/tex]
Rewrite as:
[tex]W =15\int\limits^{60}_{30} {x} \, dx[/tex]
Integrate
[tex]W =15 \frac{x^2}{2}|\limits^{60}_{30}[/tex]
This gives:
[tex]W =15 *\frac{60^2 - 30^2}{2}[/tex]
[tex]W =15 *\frac{2700}{2}[/tex]
[tex]W =15 *1350[/tex]
[tex]W =20250N-cm[/tex]
Convert to Nm
[tex]W =\frac{20250Nm}{100}[/tex]
[tex]W =202.50Nm[/tex]
