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(a) 69.3 J
The work done by the applied force is given by:
where:
F = 25.0 N is the magnitude of the applied force
d = 3.20 m is the displacement of the sled
is the angle between the direction of the force and the displacement of the sled
Substituting numbers into the formula, we find
(b) 0
The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.
(c) 69.3 J
According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:
where
is the change in kinetic energy
W is the work done
Since we already calculated W in part (a):
W = 69.3 J
We therefore know that the change in kinetic energy of the sled is equal to this value:
(d) 4.9 m/s
The change in kinetic energy of the sled can be rewritten as:
(1)
where
Kf is the final kinetic energy
Ki is the initial kinetic energy
m = 5.50 kg is the mass of the sled
u = 0 is the initial speed of the sled
v = ? is the final speed of the sled
We can calculate the variation of kinetic energy of the sled, , after it has travelled for d=3 m. Using the work-energy theorem again, we find
And substituting into (1) and re-arrangin the equation, we find