Example: Sled

Question:

A father pulls his son on a sled with a constant velocity on a level surface through a distance of 10.0 m, as illustrated. If the total mass of the sled and boy is 35 kg, and the coefficient of kinetic friction between the sled and the snow is 0.25, how much work does the father do?

Answer:

The work done on the sled is F d cosq. We are given d and q but not F. To find F, we remember that constant velocity means that sums of the applied forces are zero, + _k = 0
The various components are shown in the figure on the right and can be expressed as:
S F_x = F cosq - F_k = 0 => F cosq = F_k
S F_y = F sinq + N - mg = 0

First we remember the equation for kinetic friction:
F_k = m_k N
But careful now: N is not simply mg, because the force of the father also has a y component!

Now we just solve the two simultaneous equations for F:
F sinq + N - mg = 0
F cosq = m_k N
Result:

F = m_k mg / (cosq + m_k sinq)
= 0.25 · 35 kg · 9.8 m/s² / (cos30° + 0.25 sin30°)
= 87 N

And now that we have the strength of the force , i.e. the length of the force vector, we can calculate the work as:
W = F d cosq = 87 N · 10 m · cos30° = 750 J