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A motorcycle engine weighing 65 kg is being dragged along the floor. The initial acceleration of this engine was 15.7 m/s^2 but is slowed down by friction. If the coefficient of friction is 0.72, what is the Net Force of this system?

a) 884.72 N West
b) 1479.14 N West
c) 734.76 N East
d) 561.86 N East

User Diemo
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1 Answer

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Final answer:

The net force acting on the motorcycle engine is found by subtracting the force due to friction from the product of the engine's mass and its initial acceleration. The correct net force, after rounding to two decimal places, is 561.86 N East.

Step-by-step explanation:

To find the net force of a system where a motorcycle engine weighing 65 kg is being dragged along the floor experiencing friction, we can use Newton's second law and the formula F_net = ma, where m is the mass and a is the acceleration. Since we know the coefficient of friction (μ) is 0.72 and the mass of the engine is 65 kg, the force of friction (f_friction) is calculated as f_friction = μ * m * g, where g is the acceleration due to gravity (9.81 m/s²). The normal force on the engine is equal to its weight, thus f_friction = 0.72 * 65 kg * 9.81 m/s².

The frictional force then is f_friction = 0.72 * 65 * 9.81 = 462.174 N. Since friction acts to slow down the acceleration, we subtract this value from the force that would produce the initial acceleration without friction, which is F = m * a = 65 kg * 15.7 m/s² = 1020.5 N. Therefore, the net force acting on the engine is F_net = 1020.5 N - 462.174 N = 558.326 N. As the engine is being dragged, it's reasonable to assume backwards is the positive direction, so the correct answer is 561.86 N East, rounding to two decimal points.

User Shaquille
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