Question

A 550 kg dragster accelerates from rest to a final speed of 110 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:

$52.25\times10^4W\\699.1 hp$

Explanation:

According to the energy conversation:

ΔK=$-f_kd+W$

ΔK=$K_f-K_i ; K=1/2 mv^2$

where,

$k_i, k_f$ are initial and final kinetic energy of the system.

$v_i$= initial velocity of the system

$v_f$=final velocity of the system

W= total work done on the system

$f_k$= friction force

d= distance traveled

Given: $v_f$=110m/s

d=400m

$f_k$=1200N

$v_i$=0m/s

t=7.3s

ΔK=$-f_kd+W$

W= ΔK + $f_kd$

  =$K_f-K_i+f_kd$

  $=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500$

$P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp$