# PHYSICS PLEASE HELP ?

One block of ice slides down the (frictionless) ramp while the other falls off the end. Which block is traveling faster when it reaches the ground?

A 20.0 kg child sits at the top of a playground slide 2.0 m above the ground. Assuming very little friction on the slide, what is the child’s speed when she lands on the ground after sliding down the slide?

A machine with a 4 W motor does 20 J of useful work in 10 s. What is the machine's efficiency?

A 12 W motor is used to lift a 1.0 N object a vertical distance of 6.0 m in 2.0 s. What is the efficiency of the system?

A 1 kg model roller coaster car starts from rest at a height of 5 m. The second hill on the track has a height of 3 m. Neglecting friction, what would you expect for the speed of the car at the top of the second hill?

A 1 kg model roller coaster car starts from rest at a height of 5 m. The second hill on the track has a height of 3 m. If it comes to rest at the top of the second hill, what is the efficiency of the roller coaster?

A 100 kg roller coaster car starts from rest at a height of 140 m. The second hill on the track has a height of 95 m. If it comes to rest at the top of the second hill, what is the work done by friction on the roller coaster car?

### 2 Answers

- FiremanLv 74 months ago
One block of ice slides down the (frictionless) ramp while the other falls off the end. Which block is traveling faster when it reaches the ground?

Ans: Both will reach with the same speed,

By the law of energy conservation:

=>KE(final) = PE(initial)A 20.0 kg child sits at the top of a playground slide 2.0 m above the ground. Assuming very little friction on the slide, what is the child’s speed when she lands on the ground after sliding down the slide? Ans: By the law of energy conservation:=>KE(final) = PE(initial)=>1/2mv^2 = mgh=>v = √2gh = √[2 x 9.8 x 2] = 6.26 m/sA machine with a 4 W motor does 20 J of useful work in 10 s. What is the machine's efficiency?Ans: Output = ΔW/Δt = 20/10 = 2 Watt=>Efficiency (ƞ) = [Output/Input] x 100%=>ƞ = [2/4] x 100% = 50%A 12 W motor is used to lift a 1.0 N object a vertical distance of 6.0 m in 2.0 s. What is the efficiency of the system?Ans: Output = ΔW/Δt = (F x s)/2 = [(mg) x h]/2 = [1 x 6]/2 = 3 Watt=>Efficiency (ƞ) = [Output/Input] x 100%=>ƞ = [3/12] x 100% = 25%A 1 kg model roller coaster car starts from rest at a height of 5 m. The second hill on the track has a height of 3 m. Neglecting friction, what would you expect for the speed of the car at the top of the second hill?Ans:By the law of energy conservation:=>KE(final) = ΔPE=>1/2mv^2 = mgΔh=>v = √2gΔh = √[2 x 9.8 x 2] = 6.26 m/sA 1 kg model roller coaster car starts from rest at a height of 5 m. The second hill on the track has a height of 3 m. If it comes to rest at the top of the second hill, what is the efficiency of the roller coaster?Ans: By Efficiency (ƞ) = [Output/Input] x 100%=>ƞ = [PE(initial)/PE(final)] x 100%=>ƞ = [3/5] x 100% = 60%A 100 kg roller coaster car starts from rest at a height of 140 m. The second hill on the track has a height of 95 m. If it comes to rest at the top of the second hill, what is the work done by friction on the roller coaster car?

Ans: By the law of energy conservation:

=>PE(final) = PE(initial) + W(friction)

=>W(friction) = mgΔh

=>W(friction) = 100 x 9.8 x (140-95) = 44100 J

- Andrew SmithLv 74 months ago
There are too many questions here. It makes me suspicious. OK for 1. Gravitation al energy -> kinetic energy. IF there are no losses it doesn't matter what path has been followed the energy is the same so the final speed MUST be the same. so 2.) Ek = Pe ; 1/2 mv^2 = mgh -> v= sqrt( 2*9.8*2.0)