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

Answer:

1.9311 m/s

168.86792 N

Explanation:

v = Final velocity

m = Mass of truck = $2.4\times 10^3\ kg$

s = Displacement = 26.5 m

Work done is given by

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

The speed is 1.9311 m/s

Work done is given by

$W=Fs\\\Rightarrow F=\dfrac{W}{s}\\\Rightarrow F=\dfrac{4475}{26.5}\\\Rightarrow F=168.86792\ N$

The horizontal force is 168.86792 N