**(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