# Engineering Physics Questions and Answers – Friction and Forces – 1

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Friction and Forces – 1”.

1. The coefficient of kinetic friction is less than the coefficient of static friction.
a) True
b) False

Explanation: The value of kinetic energy is directly proportional to the normal reaction between the two surfaces. Thus the coefficient of kinetic friction is less than the coefficient of static friction.

2. The coefficient of static friction is ___________
a) Less than the coefficient of kinetic friction
b) Greater than the coefficient of limiting friction
c) Equal to the coefficient of kinetic friction
d) Equal to the tangent of the angle of friction

Explanation: The angle which the resultant of the limiting friction and the normal reaction which makes the normal reaction is called the angle of friction. But the tangent of the angle of friction is equal to the coefficient of static friction.

3. Which of the following kinetic friction is smaller?
a) Limiting friction
b) Static friction
c) Rolling friction
d) Sliding friction

Explanation: The force which comes into play when a body rolls over the surface of another body is called rolling friction. For the same magnitude of a normal reaction, rolling friction is always greater than the sliding friction.

4. A cubical block rests on an inclined plane of μ = 1/√3, determine the angle of inclination when the block just slides down the inclined plane?
a) 40°
b) 50°
c) 30°
d) 20°

Explanation: When the block just slides down the inclined plane, the angle of inclination is equal to the angle of response.
tan α = μ= 1/√3
α = 30°.

5. A mass of 4kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ= 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?
a) 0.814
b) 0.27
c) 1.5
d) 3.5

Explanation: θ = 15° is the angle of response.
Coefficient of friction, μ = tan θ = tan 15° = 0.27.
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6. A scooter weighs 120kg f. Brakes are applied so that wheels stop rolling and start skidding. Find the force of friction if the coefficient of friction is 0.4.
a) 60kg f
b) 48kg f
c) 25kg f
d) 32kg f

Explanation: Weight of the scooter = 120kg f
μ = 0.4
f = μ×weight of the scooter = 0.4×120 = 48kg f.

7. How is friction due to air reduced?
a) Streamlining
b) Lubrication
c) By using ball bearings
d) By polishing

Explanation: Friction due to air is considerably reduced by streamlining the shape pf the body moving through air. For example, jets have a streamline shape.

8. Friction can be increased by ___________
a) Using air cushion
b) Lubricants
c) Using sand
d) Using ball bearings

Explanation: By throwing sand the force of friction between the wheels and the track becomes easier. On rainy days, we throw sand on the slippery ground. This increases the friction between our feet and the ground and reduces the chance of slipping.

9. When moving along a curved path, he ___________
a) Leans inwards
b) Leans outwards
c) Is still
d) Leans sideways

Explanation: When a cyclist goes around a curved path, a centripetal force is required. The force between the tyres and the road is small to provide the necessary centripetal force. That is why a cyclist going around a curve leans inwards because the horizontal component of the normal reaction provides the necessary centripetal force.

10. A train has to negotiate a curve of radius 400m. By how much should the putter rail be raised with respect to inner rail for speed of 48 km/h? The distance between the rails is 1m.
a) 0.20m
b) 0.0454m
c) 0.45m
d) 0.020m

Explanation: h = (v2 l)/rg
h = 402/32 = 0.0454m.

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