Two Power / Work / Energy Questions

Click For Summary
The discussion revolves around calculating work and kinetic energy in two scenarios involving a car. For the first question, the work required to stop a car with a kinetic energy of 8 x 10^6 J is correctly calculated as -8 x 10^6 J, with the 10 seconds mentioned being relevant for determining power but not necessary for the work calculation. The second question involves finding the kinetic energy of a 1,500 kg car traveling at 22 m/s, which is calculated as 3.6 x 10^5 J. Participants clarify that while power can be derived from work and time, the primary focus of the first question is on work alone. The conversation emphasizes understanding the relationship between work, energy, and power in physics problems.
JayDub
Messages
30
Reaction score
0
Hey there, this is my post so I hope it does not seem like I am just asking for help without me trying to do work too.

Ok, the kinetic energy of a car is 8 x 10^6 J as it travels along a horizontal road. How much work is required to stop the car in 10s? That is the question and I am not sure about how I can solve it with Power and Work.

I know that Work = ΔE so we would have using kinetic energy

W = (1/2)mvf^2 - (1/2)mvo^2
W = (1/2)m(0m/s)^2 - 8 x 10^6 J
W = 0 - 8 x 10^6 J
W = -8 x 10^6 J

Is that just the answer? What is the 10s for? Is it just extraneous information?


Can someone just make sure I am answering this question correct

A 1,500 kg car travels at a constant speed of 22 m/s around a circular track that is 80m across. What is the kinetic energy of the car?

KE = (1/2)(m)(v)^2
KE = (1/2)(1500kg)(22m/s)^2
KE = 3.6 x 10^5 J

Thank you.
 
Physics news on Phys.org
For the first question:

Power = \frac{{Work}}{{Time}}

So you know the work, you can implement the time to determine the power.
 
Yes I realize that, however, the question asks for the work required, not the power.
 
Well then you have the answer... kinda odd to just basically give you the answer in the question though
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
961
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
1K
Replies
5
Views
3K