A spring with a tension of 200N/mm is compressed for a distance of 150mm,determine the work done if the spring is compressed a further 50mm
Given:
Tension of spring = 200 N/mm = 0.2 N/mm
Initial compression = 150 mm
Further compression = 50 mm
Work done to compress the spring further can be calculated by finding the change in potential energy stored in the spring.
Initial potential energy stored in the spring (U1) is given by:
U1 = (1/2) * k * x1^2
where,
k = spring constant = 0.2 N/mm
x1 = initial compression = 150 mm = 0.15 m
U1 = (1/2) * 0.2 * 0.15^2
U1 = 0.00225 J
Final potential energy stored in the spring (U2) is given by:
U2 = (1/2) * k * x2^2
where,
x2 = total compression after further compression = x1 + further compression
x2 = 0.15 m + 0.05 m
x2 = 0.20 m
U2 = (1/2) * 0.2 * 0.20^2
U2 = 0.0028 J
Work done to compress the spring further is equal to the change in potential energy:
Work = U2 - U1
Work = 0.0028 - 0.00225
Work = 0.00055 J
Therefore, the work done to compress the spring further by 50 mm is 0.00055 Joules.