Time to lift car with 250-W hydraulic jack

[sp00ky]

A car jack is used to lift a quarter of the weight of a 1200kg car 20 cm in 20s. Calculate the power of the car jack.

I got this answer no problem:
P = eg/t
= mgh/t
= (300kg)(9.8N/kg)(0.20m)/20.0s
= 588J/20.0s
= 29.4W

I believe this is a follow up question and I can't seem to get it. I get an answer, but it's different fromt he back of the book so I guess this answer is wrong. Maybe you can help me on this question:

A 250-W hydraulic car jack is used to do the same work on the same car as in the previous problem. How long will it take to lift the car this time?

 

[Leong]

Since the work will be the same, then
$$P_{1}*t_{1}=P_{2}*t_{2}$$

 

[wais]

To solve this problem, we can use the same formula for power: P = W/t, where P is power, W is work, and t is time. We know that the power of the car jack is 250 W, and we can calculate the work done by the jack by multiplying the force needed to lift the car (300 kg x 9.8 N/kg = 2940 N) by the distance it needs to be lifted (0.2 m). This gives us a work of 588 J. Plugging these values into the formula, we get:

250 W = 588 J/t
Solving for t, we get t = 2.35 seconds.

So, it would take approximately 2.35 seconds for the 250-W hydraulic car jack to lift the car 20 cm. This is a faster time compared to the previous problem because the car jack has a higher power output.