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Work and Power

Problems

Calculus Based Section: Variable Forces

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Problem :

A particle, starting at the origin, experiences a variable force defined by F(x) = 3x 2 , causing it to move along the x-axis. How much work is done on the particle from its starting point to x = 5 ?

We use our equation for position-dependent forces:

W = F(x)dx = 3x 2 dx = [x 3]0 5 = 53 = 125 J

The force does 125 Joules of work.

Problem :

A 2 kg mass is attached to a spring. The mass is at x = 0 when the spring is relaxed (not compressed or stretched). If the mass gets displaced from the equilibrium point ( x = 0 ) then it experiences a force from the spring described by F s = - kx , where k is a spring constant. The minus sign indicates that the force always points towards the equilibrium point, or away from the displacement of the mass.

From the equilibrium point, the mass on the spring is displaced a distance of 1 meter, then allowed to oscillate on the spring. Using our formula for work from variable forces, and the Work Energy Theorem, find the velocity of the mass when it returns to x = 0 after being initially displaced. let k = 200 N/m .

Initial Displacement of Mass

What seems like a complicated situation can be simplified using our knowledge of variable forces, and the Work-Energy Theorem. The mass is to be released from its initial displacement, and move back toward the equilibrium point, x = 0 . While it completes this journey, it experiences a force of - kx . This force does work on the mass, causing a change in its velocity. We can calculate the total work done by integration:

W = F s(x)dx = - kxdx = [ kx 2]1 0 = k(1)2 = 100 J

The spring does a total of 100 Joules of work over the trip from the initial displacement. During this time, the kinetic energy of the mass changes according to the work that was done. Since x o = 0 , we can say that:

100J = mv f 2

solving for v ,

v = = = 10 m/s

Thus the mass is traveling at a speed of 10 m/s when it crosses x = 0 .

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