Can I have some help with some physics questions?

1) A 29 kg boy climbs a 15 m rope in 45 s. What was his average power output? Use g = 10 m/s2.

Let’s use following equation to determine the power.

Power = Work ÷ time

Work = Force * distance

The force is the boy’s weight.

F = 29 * 10 = 290 N

Work = 290 * 15 = 4,350 N * m

Power = 4,350 ÷ 45 = 96⅔ watts

2) What is the minimum work needed to push a 850 -kg car 760 m up along a 9.0 ∘ incline?

The force is the component of the weight of the car that is parallel to the incline.

Force = 850 * 9.8 * sin 9 = 8,330 * sin 9

Work = Force * distance

Work = 8,330 * sin 9 * 760 = 6,330,800 * sin 9

This is approximately 9.9 * 10^5 N * m.

3) At an accident scene on a level road, investigators measure a car's skid mark to be 91 m long. It was a rainy day and the coefficient of friction was estimated to be 0.35. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.

The friction force causes the car to decelerate.

a = -0.35 * 9.8 = -3.43 m/s^2

Let’s use the following equation to determine the car’s velocity before the brakes are applied.

vf^2 = vi^2 + 2 * a * d

0 = vi^2 + 2 * -3.43 * 91

vi = √624.26

This is approximately 25 m/s. I hope this is helpful for you.

P = F*v = (29 kg)(10 m/s^2)(15 m/45 s)

= about 100 W but use a calculator.

(2) 760 m * sin(9 deg) = 119 m. That's the actual rise.

W = mg delta-h = (850 kg)(10 m/s^2)(119 m) = about 1 megajoule but use a calculator.

(3) The frictional force was (0.35)(mg), so the deceleration was 0.35g. Still working with g = 10 m/s^2 as suggested, you have

a = -3.5 m/s^2. Now use the equation

2ad = vf^2 - v0^2 =>

v0 = sqrt[(7 m/s)(91 m)] = 25 m/s.