1. Box’s mass = 2 kg, acceleration due to gravity = 9.8 m/s^{2}. Find (a) the net force which accelerates the box downward (b) magnitude of the box’s acceleration.

Solution

__Known :__

Mass (m) = 2 kg

Acceleration due to gravity (g) = 9.8 m/s^{2}

Weight (w) = m g = (2)(9.8) = 19.6 Newton

w_{x} = w sin 30 = (19.6)(0.5) = 9.8 Newton

w_{y} = w cos 30 = (19.6)(0.5√3) = 9.8√3 Newton

__Solution :__

(a) The net for__ce which accelerates the box__

Inclined plane is smooth, so there is no friction force. The only force which acts on the object is w_{x}.

∑F = w_{x}

∑F = 9.8 Newton

(b) __magnitude of the acceleration__

∑F = m a

9.8 = (2) a

a = 9.8 / 2

a = 4.9 m/s^{2}

Magnitude of the acceleration is 4.9 m/s^{2}, direction of the acceleration is downward.

2. Inclined plane is smooth so there is no friction force. Object’s mass is 3 kg, acceleration due to gravity is 9.8 m/s^{2}. Determine the magnitude of the force F if (a) object is at rest (b) object is moving downward with constant acceleration 2 m/s^{2} (c) object is moving upward with a constant acceleration of 2 m/s^{2}.

Solution

__Known :__

Mass (m) = 3 kg

Acceleration due to gravity (g) = 9.8 m/s^{2}

Weight (w) = m g = (3)(9.8) = 29.4 Newton

w_{x} = w sin 30 = (29.4)(0.5) = 14.7 Newton

w_{y }= w cos 30 = (29.4)(0.5√3) = 14.7√3 Newton

__Solution :__

(a) The magnitude of the force F if an object is at rest

Newton’s first law of motion states that if an object is at rest, the net force acts on the object is zero.

∑F = 0

F – w_{x} = 0

F = w_{x}

F = 14.7 Newton

(b) The magnitude of the force F if an object is moving downward at a constant 2 m/s^{2}

∑F = m a

w_{x} – F = m a

14.7 – F = (3)(2)

14.7 – F = 6

F = 14.7– 6

F = 8.7 Newton

(c) The magnitude of the force F if an object is moving upward at a constant 2 m/s^{2}

∑F = m a

F – w_{x} = m a

F – 14.7 = (3)(2)

F – 14.7 = 6

F = 14.7 + 6

F = 20.7 Newton

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