Question

A man does 4,475 J of work in the process of pushing his 2.40 103 kg truck from rest to a speed of v, over a distance of 26.5 m. Neglecting friction between truck and road, determine the following.

(a) the speed
(b) the horizontal force exerted on the truck

Answer 1

1.9311 m/s

168.86792 N

Step-by-step explanation:

v = Final velocity

m = Mass of truck =
`2.4* 10^3\ kg`

s = Displacement = 26.5 m

Work done is given by

`W=(1)/(2)m(v^2-u^2)\\\Rightarrow W=(1)/(2)m(v^2-0^2)\\\Rightarrow v=\sqrt{(2W)/(m)}\\\Rightarrow v=\sqrt{(2* 4475)/(2.4* 10^3)}\\\Rightarrow v=1.9311\ m/s`

The speed is 1.9311 m/s

Work done is given by

`W=Fs\\\Rightarrow F=(W)/(s)\\\Rightarrow F=(4475)/(26.5)\\\Rightarrow F=168.86792\ N`

The horizontal force is 168.86792 N