Asked by Kayla
A 1200Kg car travels at a constant speed of 26m/s for 5s up a 130m long hill. The total height of the hill is 27m & the friction force acting on the car is 2520N.
What power was used by the car in traveling up the hill?
I know that Fnet=o bc velocity is constant, therefore Fapp=2520N
I used (Fapp * d)/t so (2520 * 130)/5 to get 6.55 * 10^4
I got it incorrect... Am I supposed to do something with the height distance as well?
Thanks a bunch! :D
What power was used by the car in traveling up the hill?
I know that Fnet=o bc velocity is constant, therefore Fapp=2520N
I used (Fapp * d)/t so (2520 * 130)/5 to get 6.55 * 10^4
I got it incorrect... Am I supposed to do something with the height distance as well?
Thanks a bunch! :D
Answers
Answered by
Scott
work against friction is
... 2520 N * 130 m
change in gravitational potential is
... m g h = 1200 kg * g * 27 m
the sum is the total energy
energy divided by time (5 s) is the power
... 2520 N * 130 m
change in gravitational potential is
... m g h = 1200 kg * g * 27 m
the sum is the total energy
energy divided by time (5 s) is the power
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