An engine of a car of power 80 kW moves on a rough road with a velocity of 32 ms−¹. The forces required to bring it to rest is

A. 2.56 x 10⁶N
B. 2.50 x 10⁶N
C. 2.80 x 10³N
D. 2.50 x 10³N

Correct Answer:

Option D – 2.50 x 10³N

Explanation

In terms of dimension: power , P, = ML²T³
Force F₁ = MLT^-2; Vel, V, LT^-1
From the above, => P = F x V
i.e, ML²T^-3 = MLT^-2 x LT^-1
ML²T^-3
=> 80,000 = F x 32
∴ F = 80,000/32

= 2500N
= 2.50 x 10³N