Calculating Kinetic Energy and Work: Practice Problems and Solutions

[mark9159]

greetings...i need some help please and someone to check if i did the problems correctly.

1.) How much kinetic energy will a 5kg ball have after rolling down a 1m high incline?
My Answer: 4.43 J
If Potential Energy=mgh, then P=(5kg)(9.81m/s^2)(1m) so P=49.05 J. If this is true, then Kinetic energy = 49.05 J = 1/2mv^2. 49.05=1/2(5)v^2. Then i divided 49.05 by 2.5 (49.05/2.5) which gave me v^2=19.62. I then took the square root of 19.62 which is approximately 4.43 J.

2.) Why are no collisions perfectly inelastic?
My Answer: Almost all collisions involve some rebounding.

3.) A lever is used to raise a 300N rock a distance of 0.6 meters. You must use a force of 75N to accomplish this task. What work is done?
My Answer: 2.45 J
If work can equal mass times gravity times height, then work equals 300N times 0.6 meters, which equals 180. I then did 180 divided by 75N (the amount of force needed), coming up with the solution: 2.45 J


thank you very much for checking my work!

mark

 

[scholar]

1) After 1m, all potential energy will have been converted to kinetic energy. So your calculation should read:
Ep=Ek
mgh=Ek
Ek=5x9.81x1
Ek=49.05 J
You went the whole way and calculated the velocity, which is measured in m.s^-1. It looks as though you got confused between kinetic energy and velocity.

3) This is quite clearly incorrect. You certainly used more than 2.45 J of energy to lift a 30kg rock from the ground! Read up on levers and you should be able to get this one.

 

[Edgardo]

greetings...i need some help please and someone to check if i did the problems correctly.

1.) How much kinetic energy will a 5kg ball have after rolling down a 1m high incline?
My Answer: 4.43 J
If Potential Energy=mgh, then P=(5kg)(9.81m/s^2)(1m) so P=49.05 J. If this is true, then Kinetic energy = 49.05 J = 1/2mv^2. 49.05=1/2(5)v^2. Then i divided 49.05 by 2.5 (49.05/2.5) which gave me v^2=19.62. I then took the square root of 19.62 which is approximately 4.43 J.

EDIT! Not correct, read scholar's post. Your solution would be correct if
the question was: Calculate the velocity.

2.) Why are no collisions perfectly inelastic?
My Answer: Almost all collisions involve some rebounding.

I don't know, but your answer sounds good.

3.) A lever is used to raise a 300N rock a distance of 0.6 meters. You must use a force of 75N to accomplish this task. What work is done?
My Answer: 2.45 J
If work can equal mass times gravity times height, then work equals 300N times 0.6 meters, which equals 180. I then did 180 divided by 75N (the amount of force needed), coming up with the solution: 2.45 J

That doesn't seem correct. First of all I can see that's not correct because
of the units:
You divided 180Nm=180J by 75N which gives you 2.4m and NOT 2.45J
ALWAYS check the units, it gives you a first hint if your calculation is correct.

 

[mark9159]

so if Work= Force Times Distance, then Work= (75N)(0.6m) which equals 45 Joules...where exactly does the 300N rock come ino the equation?

the problem I am facing is that 45 J is not one of the choices i have to choose from

oh and i really thank you guys for the tip/help..im going to look up more information about levers now.

 

[Edgardo]

Mark,

I just send you a private message. I think it's just 300N times 0.6m.
The 75N is the force you need if you use a lever.

Sorry, my fault.

 

[mark9159]

oh, ok. thank you edgardo.

One more question.

What power is required to accelerate a 500kg car from zero to 18 m/s in one minute?

First i found the kinetic energy

Ek=1/2mv^2
Ek=1/2(500kg)(18m/s)^2
Ek=250(324)=81000 J

Power= Joules per second
Power= 81000 J / 60
Power= 1350 W

It takes 1350 W to accelerate a 500kg car from zero to 18 m/s in one minute.

 

[Edgardo]

Hello Mark, that seems to be correct.

Regards

Edgardo