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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?

Respuesta :

Answer:

[tex]v_f = 27.9 m/s[/tex]

Explanation:

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

[tex]F_1 = W sin\theta[/tex]

[tex]F_1 = 15000 sin15[/tex]

[tex]F_1 = 3882.3 N[/tex]

also the engine is providing the constant force to it as

[tex]F_2 = 8000 N[/tex]

now the net force along the the plane is given as

[tex]F_{net} = 8000 + 3882.3[/tex]

[tex]F = 11882.3 N[/tex]

mass of the truck is given as

[tex]m = \frac{w}{g} = 1529 kg[/tex]

now the acceleration is given as

[tex]a = \frac{F}{m}[/tex]

[tex]a = 7.77 m/s^2[/tex]

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

[tex]v_f^2 = v_i^2 + 2 a d[/tex]

[tex]v_f^2 = 0 + 2(7.77)(50)[/tex]

[tex]v_f = 27.9 m/s[/tex]

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