A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. How much work is done in compressing the spring 3 m from its equilibrium position?
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To solve this problem you can use Hooke's Law given by F = -kx and W = 0.5kx^2
F = force exerted by the spring (N)
k = spring constant (N/m)
x = distance from equilibrium (m)
First, we solve for the spring constant from the given details.
x = 0.5 m, F = 60
Solving for k we have k = F/x = 60/0.5 =120 N/m
Now we solve for for the work done on the spring, where x = 3 m
W = 0.5kx^2 = 0.5*120*3^2 = 540 J
Dimensional Analysis: N/m * m^2 = N*m = J
Good luck!
F = force exerted by the spring (N)
k = spring constant (N/m)
x = distance from equilibrium (m)
First, we solve for the spring constant from the given details.
x = 0.5 m, F = 60
Solving for k we have k = F/x = 60/0.5 =120 N/m
Now we solve for for the work done on the spring, where x = 3 m
W = 0.5kx^2 = 0.5*120*3^2 = 540 J
Dimensional Analysis: N/m * m^2 = N*m = J
Good luck!