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A 2.0 x 10^3 kg car moves down a level highway under the actions of two forces. One is a 1140 N forward force exerted on the wheels by the road. The other is a 950 N resistive force exerted on the car by the air. Use the work-kinetic energy theorem to find the speed of the car after it has moved a distance of 20.0 m, assuming the car starts from rest.

Respuesta :

Answer:

Speed of the car after it has moved a distance of 20m is 3.8 m/s

Explanation:

Mass of car is M = 2000 kg

Forces acting on car are:

Force F₁ = 1140 N forward force exerted on drive wheels by the road

Force F₂ = 950 N a resistive force

Distance moved by the car is s = 20m , after starting from rest

Work done, W = F₁ * s - F₂ * s

W = 1140 * 20 - 950 * 20 = 3800J

Assuming kinetic energy conservation is work done

W = KEf - KEi

Where KEf = final kinetic energy

KEi = initial kinetic energy which = 0 (since starting from rest)

Hence, W = 3800J = KEf = 1÷2MV²f  

Vf = final velocity = 2 * 3800 ÷ 2000 = 3.8 (m/s)²

Vf = 3.8 m/s

Speed of the car after it has moved a distance of 20m is 3.8 m/s

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