# Machine Kinematics Questions and Answers – Coefficient of Friction

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

1. A block whose mass is 650 gm is fastened to aspring constant K equals 65 N/m whose other end is fixed. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a smooth surface, and released from rest at t = 0. The maximum speed ‘S’ of the oscillating block is
a) 11 cm /sec
b) 11 m/sec
c) 11 mm/sec
d) 1.1 m/sec

Explanation: Maximum speed of the spring will be at x = 0, when total spring energy will be converted into kinematic energy of mass.
1/2 kx2 = 1/2 mv2
1/2 x 65 x(0.11)2 = 1/2 x (0.650)V2
V = 1.1 m/sec.

2. Which of the following statements regarding laws governing the friction between dry surfaces are correct?
a) The friction force is directly proportional to the normal force.
b) The friction force is dependent on the materials of the contact surfaces.
c) The friction force is independent of the area of contact.
d) all of the mentioned

Explanation: Following are the laws of solid friction :
1. The force of friction is directly proportional to the normal load between the surfaces.
2. The force of friction is independent of the area of the contact surface for a given normal load.
3. The force of friction depends upon the material of which the contact surfaces are made.
4. The force of friction is independent of the velocity of sliding of one body relative to the other body.

3. If two bodies one light and other heavy have equal kinetic energies, which one has a greater momentum
a) heavy body
b) light body
c) both have equal momentum
d) it depends on the actual velocities

Explanation: 1/2m1V12 = 1/2m2V22
v2/V1 = √m1/m2
For momentum ratio, M1/M2 = √m1/m2.

4. A heavy block of mass m is slowly placed on a conveyer belt moving with speed v. If coefficient of friction between block and the belt is μ, the block will slide on the belt through distance
a) v/μg
b) v2/√μg
c) (v/μg)2
d) v2/2μg

Explanation: Retardation due to friction force = μg
V2 = 2.μg s
s = v2/2μg.

5. A car moving with uniform acceleration cover 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. The acceleration of the car is
a) 7 m/s2
b) 50 m/s2
c) 25 m/s2
d) 10 m/s2

Explanation: s = ut + 1/2at2
at t = 5 sec, s = 450
450 = u(5) + 1/2a(25)
at t = 10 sec, s = 450 + 700 = 1150
1150 = u(10) = 1/2a(100)
a = 10 m/sec2.
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6. A particle starts with a velocity 2 m/sec and moves on a straight-line track with retardation 0.1 m/s2. The time at which the particle is 15 m from the startin g point would be
a) 10 s
b) 20 s
c) 50 s
d) 40 s

Explanation: u = 2m/sec a = 0.1 m/sec
s = ut – 1/2at2
15 = 2t – 1/2(0.1)t2
t = 30, t = 10.

7. Two particles with masses in the ratio 1 : 4 are moving with equal kinetic energies. The magnitude of their linear momentums will conform to the ratio
a) 1 : 8
b) 1 : 2
c) √2 : 1
d) √2

Explanation: 1/2m1V12 = 1/2m2V22
m1/m2 = (V2/V1)2 = 1/4
V2/V1 = 1/2
L1/L2 = m1V1/m2V2 = 1/2.

8. A stone is projected horizontally from a cliff at 10 m/sec and lands on the ground below at 20 m from the base of the cliff. Find the height of the cliff.
a) 18 m
b) 20 m
c) 22 m
d) 24 m

Explanation: h = 1/2at2
h = 1/2 x 10 x 4 = 20m.

9. A car moving with speed u can be stopped in minimum distance x when brakes are applied. If the speed becomes n times, the minimum distance over which the car can be stopped would take the value
a) x/n
b) nx
c) x/n2
d) n2x

Explanation: x = u2/2g
x’ = (nu)2/2g
x’ = n2x.

10. Ratio of the radii of planes P1 and P2 is k and ratio of the accelerations due to gravity on them is s. Ratio of escape velocities from them will be
a) ks
b) √ks
c) √k/s
d) √s/k

Explanation: v1/v2 = √g1/g2 √R2/R1
v1/v2 = √s/√k = √s/k.

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