Solved Examples: Friction | Physics for JEE Main & Advanced PDF Download

Q1: The Figure below shows blocks A and B weighing 4N and 8N, respectively and the coefficient of sliding friction between any two surfaces is 0.25. Find the force necessary to drag the block B to the left with constant velocity in all the cases when (a) A is kept over B and (b) A is held firmly over B
Sol:
In the first case, the blocks A and B move together. The friction force will be exerted on the bottom surface of B. In the second case only block B moves. The friction force will be exerted both on bottom and top surface of block B.
Let us consider free diagrams of A and B as two separate systems shown as follows:
(a) W_A = m_Ag = 4N; W_B = m_Bg = 8N
N_B = N_A + m_Bg = (m_A + m_B)g = 4+8 = 12 N
F = μ x 12 = 25 x 12 = 3N
(b) F = µ(N_A + N_B) + because B is sliding over horizontal rough surface under it and B will be sliding under A also.  

= 0.25(8 + 8) = 4N

Q2: Three blocks of masses, and are connected by inextensible strings passing over three massless pulleys as shown in the Figure. The coefficient of friction between the masses and horizontal surfaces is μ. Assume that M₁ and M₂ are sliding. Now, find
(a) Relation between accelerations a₁ , a₂ and a₃ 
(b) Tension T in the strings
Sol:
To find the constraint relation between the acceleration of the blocks, measure the distances of blocks from the stationary pulleys. Draw the FBD for each block. For M₁ and M₂ apply Newton’s second law in horizontal direction. For M₃ apply Newton’s second law in vertical direction.
(a) Forces of friction f, tension T and reaction are marked for the blocks M₁, M₂ and M₃.
Now, take the horizontal line AB as the reference line, i.e., x-axis and vertically downward as y-axis.
If x₁, x₂ and x₃ are the lengths of the strings, then x₁ + x₂ and x₃  = where L is the constant length of the string.
Now, differentiating twice, a₁ + a₂ + 2a₃ = 0
As 3 a is increasing, a₁ and a₂ are decreasing.
Thus, the constraint relation shows that a₁ + a₂ = 2a₃
(b) The equations of motion are given as follows

Q3: Masses M₁ ,M₂ and M₃ are connected by strings of negligible mass which pass over massless and frictionless pulleys P₁ and P₂ as shown in the Figure. The masses move such that the portion of the string between P₁ and P₂ is parallel to the incline and the portion of the string between P₂ and M₃ is horizontal. The masses M₂ and M₃ are 0.4 kg each and the coefficient of kinetic friction between masses and surfaces is 0.25. The inclined plane makes an angle of 37° with the horizontal, however, the mass M₁ moves with uniform velocity downwards. Now, find
(a) The tension in the horizontal portion of the string
(b) The mass M₁(g = 9.8 ms⁻² , sin 37°= 3 / 5).
Sol: Apply Newton’s first law for each of the blocks as the velocity of each block is constant.
Let T₁ be the tension between M₁ and M₂ and T₂ be the tension between M₂ and M₃.
Let μ be the coefficient of kinetic friction. Then


Q4: A block of mass m is pulled up by means of a thread up and inclined plane forming an angle α with the horizontal. The coefficient of friction is equal to µ. Find the angle β which the thread must form with the inclined plane for the tension of the thread to be minimum. Also, find the value of minimum tension.
Sol:
Draw the FBD of the block. Apply Newton’s first law along the perpendicular to the inclined plane and Newton’s second law along the inclined plane for the block.
When the body is just about to move up, the force of friction f is acting downward. If N is the normal reaction, the force of friction f is equal to μN. Further, T and mg can be resolved into rectangular components parallel and perpendicular to the inclined plane as shown in the Figure.
∴ T cos β = mg sin α +µN ....(i)
N+ T sinβ = mg cos α or N = mg cos α− T sinβ
Now, by substituting in Eq.(i), we obtain
T cos β = mg sin α+ µmg cos α− µT sinβ
or T cos β+µT sinβ = mg(sin α +µ cos α)

For T to be minimum, cos β + µ sin β should be maximum.


which is negative.
∴ For minimum T, β= tan⁻¹μ
The value of T_min can be found by writing β in terms of μ.


Q5: A body of mass 10 kg is moving with an initial speed of 20 m/s. The body stops after 5 s due to friction between body and the floor. The value of the coefficient of friction is:
(Take acceleration due to gravity g = 10 ms⁻²)
(a) 0.3
(b) 0.2
(c) 0.5
(d) 0.4
Ans: (d)
a = -μg +
∵ v = u + at
0 = 20 + (-μ x 10) x 5
50 μ = 20
μ = 2/5 = 0.4

Q6: The time taken by an object to slide down 45^∘ rough inclined plane is n times as it takes to slide down a perfectly smooth 45^∘ incline plane. The coefficient of kinetic friction between the object and the incline plane is:
(a)
(b)
(c)
(d)
Ans: (a)
smooth case:

Rough case:
a = g sin 45º - μg cos 45º


Q7: Consider a block kept on an inclined plane (incline at 45^∘) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane(µ) is equal to:
(a) 0.60
(b) 0.33
(c) 0.25
(d) 0.50
Ans: (b)






Q8: A 30 kg brick is laying on a table, not moving. What is the normal force.
Sol: 
F_n = mg
F_n = 30kg·9.8m/s²
F_n = 294N

Q9: What is the weight of a 36 kg person on earth?
Sol: W = mg
W = 36kg·9.8m/s²
W = 352.8N

Q10:  What is the weight of a 12 kg dog on the moon? (acceleration of gravity is 1.63 m/s²)
Sol: W = mg
W = 12kg·1.63m/s²
W = 19.6N

Q11: For the following problems, calculate the force of friction acting on the object. 
(i) A 10 kg rubber block sliding on a concrete floor (µ=0.65)
Sol:

(ii) A 8 kg wooden box sliding on a leather covered desk. (µ=0.40)
Sol: 

(iii) A 37 kg wooden crate sliding across a wood floor. (µ=0.20) 
Sol: 


Q14: A 248 kg object moving at 19 m/s comes to stop over a distance of 38 m. What is the coefficient of kinetic friction between the surfaces?
Sol: 
Solve for a using K3:




Q15: A hockey puck is hit on a frozen lake and starts moving with a velocity of 13.0 m/s. It travels across the ice and 6.0s later, the velocity is 7 m/s. What is the coefficient of friction between the puck and the ice?
Sol:
Solve for a using K1: v = v₀ + at

Because the net force acting on the ice is from kinetic friction, we can use the same solution from problem 1, so



Q16: A force of 36.0N acclerates a 6.0kg block at 7.0m/s² along a horiztonal surface. How large is the friction force? What is the coefficient of kinetic friction?
Sol:
F_net = 36N-f_k F_net = ma
36N-f_k = ma
36N-f_k = 6.0kg · 7.0 m/s²
36N-f_k = 42N
-f_k = 6N
f_k = μ_kF_n
μ_k = 6N/58.8N = 0.10