# PHYSICS A 1.32×103 kg car accelerates uniformly from rest to 13.1 m/s in 2.94 s. What is the work done on the car in this time interval? ?

A 1.32×103

kg car accelerates uniformly from

rest to 13.1 m/s in 2.94 s.

What is the work done on the car in this

time interval?

Answer in units of J.

017 (part 2 of 2) 10.0 points

What is the power delivered by the engine in

this time interval?

Answer in units of W.

### 3 Answers

- MyRankLv 63 months ago
Given,

Mass (m) = 1.32 x 10³kg

Speed (v) = 13.1 m/sec

Time (t) = 2.94 sec

Work done (W) =?

Power (P) =?

We know that:-

Work done (w) = Kinetic Energy (K.E) = ½mv²

= ½ x 1.32 x 10³ x (13.1)²

= ½ x 1.32 x 10³ x 171.61

= 113.26 x 10³ Joules

Now,

Power (P) = work done (W) / Time (t) = 113.26 x 10³/2.94

= 38.52 x 10³watts

∴ Power (P) = 38.52 k Watts.

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- Anonymous3 months ago
work W = KE = 1.32*10^3/2*13.1^2 = 113.26*10^3 joule

aver. power P = W/t = 113.26*10^3/2.94 = 38.5*10^3 = watt

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- Anonymous3 months ago
Calculate the KE of the car at 13.1 m/sec

Divide by the time, that gives you J/sec = W.

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