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

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Step-by-step explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m =
`m=2.70* 10^3kg`

We know that kinetic energy is given by

`KE=(1)/(2)mv^2`

So
`v=\sqrt{(2KE)/(m)}=\sqrt{(2* 4780)/(2.7* 10^3)}=1.88m/sec`

(B) We know that work done is given by

W = Fd

So
`F=(W)/(d)=(4780)/(25.5)=187.45N`