Question

A bicyclist rides 1.86 km due east, while the resistive force from the air has a magnitude of 5.12 N and points due west. The rider then turns around and rides 1.86 km due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 5.12 N and points due east. Find the work done by the resistive force during the round trip.

Answer 1

Answer:

-19.0 kJ

Explanation:

Let's keep in mind that the direction of the resistive force is always opposite to the displacement of the bicyclist.

- In the first part of the ride:

Displacement: d = 1.86 km = 1860 m

Resistive force: F = 5.12 N

Angle between the direction of the force and the displacement: $\theta=180^{\circ}$

So, the work done by the resistive force is

$W=Fdcos \theta =(5.12 N)(1860 m)(cos 180^{\circ})=-9523 J$

- Similarly, in the second part of the ride:

Displacement: d = 1.86 km = 1860 m

Resistive force: F = 5.12 N

Angle between the direction of the force and the displacement: $\theta=180^{\circ}$

So, the work done by the resistive force is

$W=Fdcos \theta =(5.12 N)(1860 m)(cos 180^{\circ})=-9523 J$

Therefore, the total work done by the resistive force during the round trip is

$W=-9523 J+(-9523 J)=-19046 J=-19.0 kJ$