Question

A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m. Neglecting friction between truck and road, determine the following. (a) the speed v (in m/s) 1.88 Correct: Your answer is correct. m/s (b) the horizontal force exerted on the truck (in N) N

Answer 1

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = $m=2.70\times 10^3kg$

We know that kinetic energy is given  by

$KE=\frac{1}{2}mv^2$

So $v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec$

(B) We know that work done is given by

W = Fd

So $F=\frac{W}{d}=\frac{4780}{25.5}=187.45N$