- #1
JayDub
- 30
- 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.
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.