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lapo4ka [179]
3 years ago
15

A 993-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen

a vertical height of 0.600 m. Friction and the drive force produced by the engine are the only two nonconservative forces present. Friction does -2870 J of work. How much work does the engine do
Engineering
1 answer:
tino4ka555 [31]3 years ago
7 0

Answer:

13177.34 J

Explanation:

Work done = force × distance

work done by the engine = kinetic energy + potential energy + work done friction

kinetic energy due to the car's speed = 1/2mv² = 4468.5 J

potential energy due to the height = mgh = 993 kg × 9.8 m/s² × 0.6 m = 5838.84 J

work done by friction = 2870 J

work done by engine = 5838.84 J + 2870 J + 4468.5 J = 13177.34 J

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Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Di
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0.556 Watts

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Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=\frac{2\times (s-ut)}{t^2}\\\Rightarrow a=\frac{2\times (5-0)}{29^2}=\frac{10}{481}

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