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Alenkinab [10]
2 years ago
7

A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.

Neglecting friction between truck and road, determine the following. (a) the speed v (in m/s) 1.88 Correct: Your answer is correct. m/s (b) the horizontal force exerted on the truck (in N) N
Physics
1 answer:
wolverine [178]2 years ago
4 0

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = m=2.70\times 10^3kg

We know that kinetic energy is given  by

KE=\frac{1}{2}mv^2

So v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec

(B) We know that work done is given by

W = Fd

So F=\frac{W}{d}=\frac{4780}{25.5}=187.45N

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