# Motion on the inclined plane without the friction force – application of Newton’s law of motion problems and solutions

1. Box’s mass = 2 kg, acceleration due to gravity = 9.8 m/s2. 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/s2

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

wx = w sin 30 = (19.6)(0.5) = 9.8 Newton

wy = w cos 30 = (19.6)(0.5√3) = 9.8√3 Newton

Solution :

(a) The net force which accelerates the box

Inclined plane is smooth, so there is no friction force. The only force which acts on the object is wx.

F = wx

F = 9.8 Newton

(b) magnitude of the acceleration

F = m a

9.8 = (2) a

a = 9.8 / 2

a = 4.9 m/s2

Magnitude of the acceleration is 4.9 m/s2, direction of the acceleration is downward.

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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/s2. Determine the magnitude of the force F if (a) object is at rest (b) object is moving downward with constant acceleration 2 m/s2 (c) object is moving upward with a constant acceleration of 2 m/s2. 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

wx = w sin 30 = (29.4)(0.5) = 14.7 Newton

wy = 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 – wx = 0

F = wx

F = 14.7 Newton

(b) The magnitude of the force F if an object is moving downward at a constant 2 m/s2

F = m a

wx – 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/s2

F = m a

F – wx = m a

F – 14.7 = (3)(2)

F – 14.7 = 6

F = 14.7 + 6

F = 20.7 Newton

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