i need help with some physics questions and my coursework is due in tomorrow. can anyone help?

1) 100km/h * 1m/s / 3.6km/h = 27.8 m/s

a) work = F * d = F * (v * t) = 500N * (27.8m/s * 20*60s) = 1.67e7 J = 16.7 MJ

b) power = F * v = 500N * 27.8m/s = 13 900 W = 13.9 kW

2) If applied force = weight, then the acceleration = g.

To be dissipated over 10 seconds at this rate, the initial velocity must have been

v = g*t = 9.8m/s² * 10s = 98 m/s

so the initial KE = ½mv² = ½ * 20kg * (98m/s)² = 96 000 J

which is equal to the work done, and the power

P = work / time = 96000J / 10s = 9600 W = 9.6 kW

3) Depends on the friction force.

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Since the car and trailer are moving at a constant velocity, the net force is 0 N. This means the applied force is equal to the friction force. To the work, we need to determine the distance they move in minutes. Let’s convert the velocity to meters per minute.

v = 100,000 ÷ 60 = 1,666⅔

To determine the distance, multiply by 20 minutes.

d = 20 * 1,666⅔ = 33,333⅓ meters

Work = Force * distance

Work = 500 * 33,333⅓ = 16,666,666⅔ N * m

Power = Work ÷ time in second

t = 20 * 60 = 1200 seconds.

Power = 16,666,666⅔ ÷ 1200

The power is approximately 138,889 watts.

2)Calculate the work done and the power dissipated in a moving body along a horizontal surface 15 metres at a constant velocity, When the only applied force is equal to the weight of the body the movement lasts 10 seconds and the mass of the body is 20 kilograms

This is similar to the problem above. “When a object is moving on a horizontal surface at a constant velocity, the applied force is equal to the friction force”.

Force = 20 * 9.8 = 196 N

Work = 196 * 15 = 2940 N * m

Power = 2940 ÷ 10 = 294 watts

3) Calculate the work done and the power dissipated for a mass of weight 500n to be horizontally moved at a constant velocity for 160 metres when the whole movement lasts 20 seconds.

This is similar to the problem above.

W = F * d

P = W ÷ t

I hope this helps you to understand how to solve these types of problems