Question

A 15,000 N truck starts from rest and moves down a 15∘ hill with the engine providing a 8,000 N force in the direction of the motion. Assume the rolling friction force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill?

Answer 1

Answer:

$v_f = 27.9 m/s$

Explanation:

Component of the weight of the truck along the inclined plane is given as

$F_1 = W sin\theta$

$F_1 = 15000 sin15$

$F_1 = 3882.3 N$

also the engine is providing the constant force to it as

$F_2 = 8000 N$

now the net force along the the plane is given as

$F_{net} = 8000 + 3882.3$

$F = 11882.3 N$

mass of the truck is given as

$m = \frac{w}{g} = 1529 kg$

now the acceleration is given as

$a = \frac{F}{m}$

$a = 7.77 m/s^2$

now the speed of the truck after travelling distance of d = 50 m is given as

$v_f^2 = v_i^2 + 2 a d$

$v_f^2 = 0 + 2(7.77)(50)$

$v_f = 27.9 m/s$