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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:

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.

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:

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.