7. A 1.50 x 103 kg car accelerates uniformly from rest to 10.0 m/s in 3.00 s.

a. What is the work done on the car in this time interval?

b. What is the power delivered by the engine in this time interval?

Answers

Answer 1

Work done = force*distance = mass*acceleration*distance

acceleration = 10/3 = 3.33 m/s^2

distance = ((0+10)/2)*3 = 15m

work done = 1500*3.33*15 = 74900 J

Power = workdone/time = 74900/3 = 25000 W

Answer 2

a. The work done on the car in this time interval is 75000 J.

b. And, the power delivered by the engine in this time interval is 25,000 watt.

Given that,

A 1.50 x 103 kg car accelerates uniformly from rest to 10.0 m/s in 3.00 s.

Based on the above information, the calculation is as follows:

(a)

We know that

[tex]w = \frac{1}{2} \times m (v^2 - u^2)[/tex]

Here

Mass = [tex]1.50 \times 103 kg[/tex]

v= 10 m / s

u= 0 m / s

Now place it to the above formula

So, Work done is  = 75000 J

(b)

[tex]Power P = W \div t\\\\= 75000 \div 3[/tex]

= 25000 Watt

Therefore the above are the answers.

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