Friction Test

Result

Friction Test - 4

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Question 1
4 / -1

On the horizontal surface of a truck, a block of mass 1 kg is placed (µ = 0.6) and truck is moving with an acceleration of 5 m/s², then the frictional force on the block will be—

A
5 N

B
6 N

C
5.88 N

D
8 N

Solution

F = µR = µ mg = 0.6 × 1× 9.8 = 5. 88 N
Question 2
4 / -1

A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless

pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (given g = 10 m/s²)—

A
100 m/s²

B
3 m/s²

C
10 m/s²

D
30 m/s²

Solution

For body A, T = M₁ a = 7 a

For body B, M₂ g _T = M₂ a

3 g-7 a = 3 a

10 a = 3 g
Question 3
4 / -1

A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (vertically). If the acceleration of the system is (g/8), then the ratio of masses is—

A
8 : 1

B
9 : 7

C
4 : 3

D
5 : 3

Solution

In the given system,
Question 4
4 / -1

A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping through these motion). The direction of the frictional force acting on cylinder are

A
Up the incline while ascending and down the incline while descending

B
Up the incline while ascending as well as descending

C
Down the incline while ascending and up the incline while descending

D
Down the incline while ascending as well as descending

Solution

While ascending the cylinder tries to pull the ground backwards while moving forward, thus frictional force acts upwards. During descending the frictional force again acts upwards. If friction were absent the body would slide down. Thus the frictional force in this case is responsible for rolling motion of the cylinder
Question 5
4 / -1

A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively.

A
g,g

B
g−a,g−a

C
g−a,g

D
a,g

Solution

Lift is accelerating downwards, so acceleration of lift w.r.t. ground  a_lg = a

Acceleration of ball w.r.t. ground  a_bg = g

Acceleration of ball w.r.t. lift   a_bl = a_bg − a_lg = g − a
Question 6
4 / -1

A man weighs 80kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s². What would be the reading on the scale.

A
1200

B
500

C
0

D
144

Solution

Mass of man m = 80 kg

Acceleration of the lift moving upwards a=5 m/s²

Acceleration due to gravity g=10 m/s²

Normal reaction acting on the man N=m(a+g)

∴ N=80(5+10)=1200 Newton
Question 7
4 / -1

In a gravity free space, a man of mass M standing at a height h above the floor, throws a ball towards the ground. When the ball reaches the floor, the distance of the man above the floor will be:

A

B
h(2−m/M)

C
2h

D
a function of m,M,h and u

Solution

In the gravity free space, lets consider man and ball be our system.

For a system, the center of mass remains same if there are no external forces on the system i.e. internal forces cannot change the center of mass of system.

Therefore, net change in center of mass =0

After the ball is thrown let the man moves a distance x in opposite direction.

Let, ↑ direction be positive and viceversa

So, Mx + m (−h) = 0 ⇒ x = hm/M

Final distance from floor = initial distance + change after ball is thrown = h+ hm/M = h (1+m/M)
Question 8
4 / -1

A car starts from rest to cover distance s. The coefficient of kinetic friction between the road and tire is μ. The minimum time in which the car cover the distance s is proportional to

A
μ

B

C
1/μ

D
1/√μ

Solution
Question 9
4 / -1

A car starts from rest and moves on a surface in which the coefficient of friction between the road and the tyres increases linearly with distance (x). If the car moves with the maximum possible acceleration, the Kinetic energy (E) of the car will depend on x as:

A
E ∝ 1/x²

B
E ∝ 1/x

C
E ∝ x

D
E ∝ x²

Solution

Support car moves a distance dx at a distance x.

μ = kx

f_r = kx(mg)

Work done by friction = K.E(E)

E = f_rdx = mgkxdx

E ∝ x²
Question 10
4 / -1

In the Figure, the ball A is released from rest when the spring is at its natural length. For the block B, of mass M to leave contact with the ground at some stage, the minimum mass of A must be:

A
2 M

B
M

C
M/2

D
A function of M and the force constant of the spring.