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