AN INTERMEDIATE LEVEL PHYSICS QUESTION. HELP!?
A 1050 KG car drives up a hill that is 16.2 high. During the drive, two nonconservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car's engine. The work done by friction is −3.01×105 j ; the work done by the engine is 6.64×105 j .
Find the change in the car's kinetic energy from the bottom of the hill to the top of the hill.
please help me it is due in two hours :(
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From conservation of energy
(Change in KE) + (Change in PE) = (WD by friction) + (WD by engine)
(Change in KE) + (1050 X 9.81 x 16.2) = (-3.01 x 105) + (6.64 x 105)
(Change in KE) + 166868 = 3.63 x 105
(Change in KE) = 196,000 J to 3 significant figures
W(net) = W(total) - W(friction)
=>W(net) = 6.64 x 10^5 - 3.01 x 10^5 = 3.63 x 10^5 j
By work energy relation:-
=>W(net) = PE + KE
=>3.63 x 10^5 = mgh + 1/2mv^2
=>3.63 x 10^5 = 1050 x 9.8 x 16.2 + 1/2 x 1050 x v^2
=>v = â373.91
=>v = 19.33 m/s
The change in kinetic energy is equal but opposite in sign to the change in potential energy that occurred when the car went from the bottom to the top of the hill.
ÎE = -(m)(g)(h) = -(1050 kg)(9.8 m/s²)(16.2 m) = 1.666980 à 10^5 J