Chapter 1: Q74P (page 199)

You and your bicycle have combined mass 80.0 kg. When you reach the base of a bridge, you are traveling along the road at 5.00 m/s (Fig. P6.74). At the top of the bridge, you have climbed a vertical distance of 5.20 m and slowed to 1.50 m/s. Ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge?(b) How much work have you done with the force you apply to the pedals?

Short Answer

Expert verified

The work done on the person and the bicycle is $- 910 J$ .
The work done by the person on the pedals is $3.17 \times 10^{3} J$ .

Step by step solution

Conservation of Energy

According to the law of conservation of energy, in a process the total energy remains conserved, it transforms from one form of energy to another form of energy.

Given data

The combined mass of the person and bicycle is, $m = 80 kg$

The initial velocity of the bicycle is, $v = 5.0 m / s$

The vertical distance climbed by the bicycle is, $h = 5.20 m$

The final velocity of the bicycle is, $v_{f} = 150 m / s$

(a)Find the total work done on the person and the bicycle

Work done is,

$W = K_{2} - K_{1} = \frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}^{2} = \frac{1}{2} (80 kg) [{(1.50 m / s)}^{2} - {(5.0 m / s)}^{2}] = - 910 J$

Hence, the work done on the person and the bicycle is $- 910 J$ .

(b) Find the work done by the person on the pedals

Work done is,

$W_{tot} = (K_{2} - K_{1}) + mgh = \frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}^{2} + mgh = \frac{1}{2} (80 kg) [{(1.50 m / s)}^{2} - {(5.0 m / s)}^{2}] + (80 kg) (9.8 m / s^{2}) (5.20 m) = 3.17 \times 10^{3} J$