Question

A man does 4,465 J of work in the process of pushing his 2.90 103 kg truck from rest to a speed of v, over a distance of 25.0 m. Neglecting friction between truck and road, determine the following.(a) the speed vm/s
(b) the horizontal force exerted on the truck
N

Answer 1

Answer:

(a)The final speed of the truck "v"  is 1.76 m/s

(b) The horizontal force exerted on the truck is 178.64 N

Explanation:

Given;

work done by the man, W = 4,465 J

mass of the truck, m = 2.9 x 10³ kg

initial speed of the truck, u = 0

distance the truck was pushed to, d = 25 m

final speed of the truck, = v

The work done in pushing the truck is calculated as;

W = F x d

where;

F is the horizontal force applied

d is the distance in which the truck was pushed to

W = ma x d

$a = \frac{W}{md} \\\\a = \frac{4465}{2.9 \times 10^3 \ \times 25} \\\\a = 0.0616 \ m/s^2$

(a)The final speed of the truck "v"  is calculated as;

v²= u² + 2ad

v² = 0 + 2(0.0616 x 25)

v² = 3.08

v = √3.08

v = 1.76 m/s

(b) the horizontal force exerted on the truck

F = ma

F = (2.9 x 10³) (0.0616)

F = 178.64 N