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?

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

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,

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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