Question

The swimmer shown in Figure 7.44 exerts an average horizontal backward force of 80.0 N with his arm during each 1.80 m long stroke.

(a) What is his work output in each stroke?

(b) Calculate the power output of his arms if he does 120 strokes per minute.

Figure 7.44

Short answer

(a) The work output in each stroke is\(144{\rm{ J}}\).

(b) The power output of his arms is \(288{\rm{ W}}\).

Step-by-step solution

Work done and Power

When a force acts on a body it causes some displacement, then the work done is,

\(W = Fd\)

Here, F is the force acting on the body, and d is the displacement.

Work output of each stock

(a)

The work output of each stock is,

\({W_1} = Fd\)

Here, F is the average horizontal force\(\left( {F = 80.0{\rm{ N}}} \right)\), and d is the distance covered in a stock\(\left( {d = 1.80{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned} {W_1} &= \left( {80.0{\rm{ N}}} \right) \times \left( {1.80{\rm{ m}}} \right)\\ &= 144{\rm{ J}}\end{aligned}\)

Therefore, the required work output in each stroke is \(144{\rm{ J}}\).

The power output of his arms

(b)

If the swimmer can perform\(120{\rm{ stokes}}/{\rm{min}}\), the power output of the swimmer is,

\(P = {W_1} \times {\rm{Rate}}\)

Putting all known values,

\(\begin{aligned} P &= \left( {144{\rm{ J}}} \right) \times \left( {120/{\rm{min}}} \right)\\ &= \left( {144{\rm{ J}}} \right) \times \left( {120/{\rm{min}}} \right) \times \left( {\frac{{1{\rm{ min}}}}{{60{\rm{ sec}}}}} \right)\\ &= 288{\rm{ W}}\end{aligned}\)

Therefore, the required power output of his arms is\(288{\rm{ W}}\).