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

`52.25*10^4W\\699.1 hp`

Step-by-step 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\\=(1)/(2) * 550*110^2 - (1)/(2) * 550*0^2+ (1200*400)\\=3807500`

`P=(W)/(t) =(3807500)/(7.3) \\P=52.15 *10^4w\\P=(52.15 *10^4)/(746) \\P=699.1 hp`