A 2000 kg car moves down a level highway under the actions of two forces. one is a 1060 n forward force exerted on the drive whe
els by the road; the other is a 1010 n resistive force. use the work-energy theorem to find the speed of the car after it has moved a distance of 21 m, assuming it starts from rest.
1 answer:
Refer to the diagram shown below.
The net force acting on the vehicle is
F - R = 1060 -1010 = 50 N
The distance traveled is 21 m. Because the force is constant, the work done is
W = (50 N)*(21 m) = 1050 J
Assume that energy is not dissipated by air resistance or otherwise.
Conservation of energy requires that W = KE, where KE is the kinetic energy of the vehicle.
The KE is
KE = (1/2)*(2000 kg)*(v m/s)² = 1000v² J
Equate KE and W to obtain
1000v² = 1050
v² = 1.05
v = 1.025 m/s
Answer: 1.025 m/s
The net force acting on the vehicle is
F - R = 1060 -1010 = 50 N
The distance traveled is 21 m. Because the force is constant, the work done is
W = (50 N)*(21 m) = 1050 J
Assume that energy is not dissipated by air resistance or otherwise.
Conservation of energy requires that W = KE, where KE is the kinetic energy of the vehicle.
The KE is
KE = (1/2)*(2000 kg)*(v m/s)² = 1000v² J
Equate KE and W to obtain
1000v² = 1050
v² = 1.05
v = 1.025 m/s
Answer: 1.025 m/s
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