# Motion on the rough inclined plane with the friction force – application of Newton’s law of motion problems and solutions

1. Object’s mass = 2 kg, acceleration due to gravity = 9.8 m/s^{2}, coefficient of the static friction = 0.2, coefficient of the kinetic friction = 0.1. Is the object at rest or accelerating? If the object is accelerated, find (a) the net force (b) magnitude and direction of the box’s acceleration!

Solution

__Known :__

Mass (m) = 2 kg

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

Coefficient of the static friction (μ_{s}) = 0.2

Coefficient of the kinetic friction (μ_{k}) = 0.1

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

The horizontal component of the weight (w_{x}) = w sin 30^{o} = (19.6)(0.5) = 9.8 Newton

The vertical component of th weight (w_{y}) = w cos 30^{o} = (19.6)(0.5√3) = 9.8√3 Newton

The normal force (N) = w_{y} = 9.8√3 Newton

Force of the static friction (f_{s}) = (0.2)(9.8√3) = 1.96√3 Newton = 3.39 Newton

Force of the kinetic friction (f_{k}) = (0.1)(9.8√3) = 0.98√3 Newton = 1.69 Newton

__Solution :__

Object is at rest if w_{x} < f_{s}, object is moving down if w_{x} > f_{s}.

w_{x} = 9.8 Newton and f_{s }= 3.39 Newton.

__(a) the net force__

∑F = w_{x }– f_{k} = 9.8 – 1.69 = 8.11 Newton

__(b) magnitude and direction of the acceleration__

∑F = m a

8.11 = (2) a

a = 4.05

Magnitude of the acceleration = 4.05 m/s^{2 }and direction of the acceleration = downward.

2. Object’s mass = 4 kg, acceleration due to gravity = 9,8 m/s^{2}. Coefficient of the kinetic friction = 0.2 and coefficient of the static friction = 0.4. Magnitude of the force F = 40 Newton. The object is at rest or slides down ? If the object slides down, find (a) the net force (b) magnitude and direction of the acceleration!

Solution

__Known :__

Mass (m) = 4 kg

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

The coefficient of the static friction (μ_{s}) = 0.4

The coefficient of the kinetic friction (μ_{k}) = 0.2

Weight (w) = m g = (4)(9.8) = 39.2 Newton

The horizontal component of the weight (w_{x}) = w sin 30^{o} = (39.2)(0.5) = 19.6 Newton

The vertical component of the weight (w_{y}) = w cos 30^{o} = (392)(0..5√3) = 19.6√3 Newton

The normal force (N) = w_{y} = 19.6√3 Newton = 33.95 Newton

the static friction force (f_{s}) = μ_{s }N = (0,4)(33.95) = 13.58 Newton

The kinetic friction force (f_{k}) = μ_{k }N = (0.2)(33.95) = 6.79 Newton

F = 40 Newton

__Solution :__

The object slides down if F < w_{x} + f_{s}. The object slides up if F > w_{x} + f_{s}.

F = 40 Newton, w_{x} = 19.6 Newton and f_{s }= 13.58 Newton.

F is greater than w_{x} + f_{s }so the object slides up.

__(a) The net force__

∑F = F – w_{x} – f_{k} = 40 – 19.6 – 6.79 = 13.61 Newton

__(b) The magnitude and direction of the acceleration __

∑F = m a

6.4 = (4) a

a = 1.6

The magnitude of the acceleration is 1.6 m/s^{2 }and direction of the acceleration is upward.

### Ebook PDF motion on rough incline plane with friction force sample problems with solutions

1 file(s) 68.37 KB- Mass and weight
- Normal force
- Newton’s second law of motion
- Friction force
- Motion on the horizontal surface without friction force
- The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
- Motion on the inclined plane without friction force
- Motion on the rough inclined plane with the friction force
- Motion in an elevator
- The motion of bodies connected by cord and pulley
- Two bodies with the same magnitude of accelerations
- Rounding a flat curve – dynamics of circular motion
- Rounding a banked curve – dynamics of circular motion
- Uniform motion in a horizontal circle
- Centripetal force in uniform circular motion