Question

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 n. what is its average power output in watts and horsepower if this takes 7.30 s?

Answer 1

Answer:

276712.33 Watt , 371 h p

Explanation:

m = 500 kg, u = 0, v = 100 m/s, s = 400 m, t = 7.30 s

frictional force = 1200 N

Use third equation of motion

v^2 = u^2 + 2 as

100 x 100 = 0 + 2 x a x 400

a = 12.5 m/s^2

Applied force, F = m x a = 500 x 12.5 = 6250 N

Net Force, Fnet = applied force - frictional force = 6250 - 1200 = 5050 N

Power, P = Work / time = Net force x distance / time

P = 5050 x 400 / 7.30 =  276712.33 Watt

1 hp = 746 watt

So, P = 276712.33 / 746 = 371 h p

Answer 2

In order to accelerate the dragster at a speed $v_f = 100 m/s$, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$:
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$