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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.

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 1

Solution

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 2

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.

See also  Springs in series and parallel – problems and solutions

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.

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 3

Solution

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 4

Known :

Mass (m) = 3 kg

Acceleration due to gravity (g) = 9.8 m/s2

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

See also  Freely falling objects – problems and solutions

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  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

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