A 15 kg mass, initially at rest, is acted on by a single, constant, external force, which does 3.0 x 10³ J of work on the mass. There is no friction.

a) Find the final speed of the mass.

b) Assume the mass moved a distance of 20 m while the force was acting; what was the magnitude of the force?

c) Find the length of time that the force acted.

a) The work done is the change in the kinetic energy:

W = K

   = ½ mv² - ½ mv₀²

   = ½ mv² - 0

or

v² = 2W/m

v = [2W/m]^1/2 = [2(3.0 x 10³J)/(15 kg)]^1/2

   = 20 m/s

b) The force is constant and parallel to the displacement (since there is only one force), thus

W = Fx

or

F = W/x

   = (3.0 x 10³J)/(20 m) = 150 N

c) Using F = ma we have

a = F/m = 150 N/15 kg = 10 m/s²

The motion is under constant acceleration, so

v = v₀ + a t

   = at

t = v/a = (20 m/s)/(10 m/s²) = 2.0 s

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