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SSSSS [86.1K]
3 years ago
13

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 mo

tion. 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?
Physics
1 answer:
GenaCL600 [577]3 years ago
7 0

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

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Answer:

t=2.5\times 10^{-14}\ s

Explanation:

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Now by putting the value

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