# Physics Questions and Answers – Friction

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Friction”.

1. Static coefficient of friction is greater than the kinetic coefficient of friction.
a) True
b) False

Explanation: Since the value of static frictional force is greater than the kinetic frictional force, dividing both values with constant normal reaction force yields a greater value of the static coefficient of friction.

2. What are the types of kinetic friction?
a) Sliding friction, rolling friction and adhesive friction
b) Sliding friction and rolling friction
c) Rolling friction and adhesive friction
d) Sliding friction and adhesive friction

Explanation: Kinetic friction is of two types; sliding and rolling. Rolling friction is less than sliding friction. This is the reason why rolling a body is always easier than sliding a body.

3. The maximum value of static friction when the body is at the verge of starting motion is known as _____
a) Static friction
b) Limiting friction
c) Impending motion
d) Angle of repose

Explanation: The value of limiting friction can be found by equating it to the product of the static frictional coefficient and normal reaction force of the body.

4. For a body of mass “m” on a rough inclined plane with a constant but arbitrary coefficient of friction as shown in the figure, what is the angle of repose if the net downward force is donated as “F”. Let “g” be the acceleration due to gravity and “u” be the coefficient of friction. a) sin θ
b) cos θ
c) tan θ
d) cot θ

Explanation: To find the angle of repose;
Net downward force = frictional force
F=f
mg (sin θ) = uR
= u x (mgcos θ)
u = tan θ.

5. For a body of mass “m” on a rough inclined plane with a constant but arbitrary coefficient of friction as shown in the figure, what is the acceleration on the inclined plane if the net downward force is donated as “F”. Let “g” be the acceleration due to gravity and “u” be the coefficient of friction. a) g x sin θ
b) g x cos θ
c) g (sin θ – u cos θ)
d) g (cos θ – u sin θ)

Explanation: To find the acceleration on the inclined plane;
ma = mg sin θ – u mg cos θ
a = g (sin θ – u cos θ).

6. _____ is known as the motion that would take place under the applied force if friction were absent.
a) Impending motion
b) Constrained motion
c) Unconstrained motion
d) Linear motion

Explanation: Impending motion is known as the motion that would take place under the applied force if friction were absent. The motion which can’t proceed arbitrarily in any manner is called constrained motion. Motion without any restriction is known as an unconstrained motion. One dimensional motion along a straight line is known as linear motion.

7. If a box is lying on the floor of a wagon with a coefficient of friction 0.2, what is the maximum acceleration of the wagon for which the box would remain stationary? (Let g = 9.81 m/s2)
a) 1.96 m/s2
b) 2 m/s2
c) 3.92 m/s2
d) 4 m/s2

Explanation: Let the mass of the box be “m”
ma = u x mg
a = u x g
a = 0.2 x 9.81
= 1.96 m/s2.

8. Limiting friction of a body depends on _____
a) Area of contact of surfaces
b)The volume of the smaller body on larger surface
c) Nature of surfaces
d) The periphery of the contact surfaces

Explanation: Since limiting friction depends only of the coefficient of friction of the surfaces, it, therefore, depends only of the nature of the surface, i.e., smoothness or roughness.

9. Consider a block of mass 9.81kg on an inclined surface of the coefficient of friction 0.15 and angle of inclination 30 degrees. What will be its acceleration on the inclined plane if acceleration due to gravity is 9.81 m/s2?
a) 3.63 m/s2
b) 4.63 m/s2
c) 9.81 m/s2
d) 10 m/s2

Explanation: a = g (sin θ – u cos θ)
a = 9.81 (sin(30) – 0.15 cos(30))
= 9.81 x (0.5 – 0.13)
= 3.63 m/s2.

10. The value of the frictional force for a body is always constant when it is static and does not vary with applied force.
a) True
b) False 