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Numerical problems based on friction

Numerical Problems based on concept of friction

Numerical Problems Based On Concept of Friction

Question 1: An object of mass 2 kg resting on the floor. The coefficient of static friction between the object and the floor is µ = 0.8. What force must be applied on the object to move it?
Answer:15.68 N

Solution: Given : m=2kg,µ=0.8
Friction force = µmg=0.8*2*9.8=15.68 N

Question 2: An object of mass 50 kg at rest on the floor. A Force of 5 N is applied on the object but it does not move. What is the frictional force that acts on the object?
Answer:5 N

Solution: Applied force =5N
Object is at rest.
Friction force =5N

Question 3: : A child is pulling a box of mass 10 Kg. calculate the frictional force if the coefficient of friction μ is 0.3?
Answer:29.4 N

Solution: Given : m=10kg,µ=0.3
frictional force=µmg=0.3*10*9.8=29.4 N

Question 4: : Rita of mass 40 Kg is slipping on the frost. If the coefficient of friction acting is 0.45. Find the frictional force acting between her and the frost layer?
Answer:176.4 N

Solution: Given : m=40kg,µ=0.45
frictional force =µmg =0.45*40*9.8=176.4 N

Question 5: :Force of 200N is exerted at an angle of 30° on the box kept of mass 30kg on the floor with the coefficient of static friction of 0.2. Find the friction force?(take g=10 m/s²)
Answer:40 N

Solution: Given : F = 200N at 30° with horizontal
Fx = 200*Cos30 = 173.2 N.
Fy = 200*sin30 = 100 N.
Wt. of the object =M*g = 30kg * 9.8 = 294 N acting downwards
Reaction of surface on box N = mg-Fy =(30*10)-100=200N
Force of kinetic friction(Fk) = µ*N = 0.2*200=40N

Question 6: :A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 ms-2, calculate the frictional force acting on the block.
Answer:5 N

Solution:

Limiting friction, f = μmg=0.6×1×9.8==5.88 N
Applied force, F = ma =1× 5=5N
As Applied force < Limiting frictional force,
so force of friction = 5 N
Question 7: Force of 50N is exerted horizontally on the box of 50Kg kept on the floor with the coefficient of static friction of 0.3. Find the friction force.
Answer:50 N

Solution: Limiting friction, f = μmg=0.3×50×10==150 N
Applied force, F = 50N
As Applied force < Limiting frictional force,
so force of friction = 50 N

Question 8: A block of Mass M is moving with a velocity v on straight surface. What is the shortest distance and shortest time in which the block can be stopped if is coefficient of friction µ.
Answer: s =v²/2(µg),t=u/µg

Solution: In this problem frictional force will act as retarding force.
Frictional force acting on block mass M = µMg
Retardation =-µMg/M=-µg
Initial velocity (u)=v ,Final velocity(v) =0
Let shortest distance that object will cover= s.
Accoding to equation of motion
v²=u²+2as
0=v²+2(-µg)s
v²=2(µg)s
s =v²/2(µg)
shortest distance that object will cover= s = v²/2(µg)
Let after time t object will come at rest.
Accoding to equation of motion
v=u+at
0=u+(-µg)t
t=u/µg

Question 9: A block of mass M=5kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F=40N is applied in horizontal direction, the acceleration of the block will be then (g=10ms2).
Answer: 6 m/s²

Solution: Let acceleration of block be a .
Force applied on the block F= 40N
Friction force acting on the block f= 0.2*5*10=10N
Friction force is opposite of applied force.
Net force acting on the block=40-10=30N
Therefore a=30/5= 6 m/s²

Question 10: Aman is hauling a toy car of mass 4 kg which was at rest earlier on the floor. If 50 N is the value of the static frictional force, calculate the friction coefficient?
Answer: 1.282

Solution: static friction = µmg
50=µ*4*9.8
µ=50/4*9.8=1.28

Question 11: : Two bodies of masses 7 kg and 5 kg are connected by a light string passing over a smooth pulley at the edge of the table as shown in the figure. The coefficient of static friction between the surfaces (body and table) is 0.9.(a) Will the mass m1 = 7 kg on the surface move? (b) If not what value of m2 should be used so that mass 7 kg begins to slide on the table?

Answer: (a)m1 will not move.(b)m2>6.3

Solution: Let tension in string T. To move m1,
Tention T> µm1g (Limiting friction on m1 mass) or T>0.9*7*10
T>63
Calculation of T :
T=m2g
T=5*10
T=50
Here we see T< Limiting friction on m1 mass.
Hence m1 will not move.
To move m1 :
T>63
m2g>63
m2>6.3 kg

Question 12: : A block of mass M = 10 kg is sitting on a surface inclined at angle θ = 45°. Given that the coefficient of static friction is μs = 0.5 between block and surface, what is the minimum force F necessary to prevent slipping? What is the maximum force F that can be exerted without causing the block to slip?

Answer: Fmin = =34.20 N , Fmax = 103.3 N

Solution: The minimum force required to prevent slipping is equal to the minimum force that will prevent the block from sliding down .
Fmin = 10gsin(45°)−10gcos(45°)×0.5
Fmin = 10*9.8*0.707−10*9.8*0.707*0.5=34.20 N
The maximum force required on block to motion in upward direction
Fmax = 10gsin(45°)+10gcos(45°)×0.5
Fmax = 10*9.8*0.707+10*9.8*0.707*0.5= 103.3 N

Question 13: A block weighing 200 N is pushed along a surface. If it takes 80 N to get the block moving and 40 N to keep the block moving at a constant velocity, what are the coefficients of friction μs and μk?
Answer: μs = 0.4 , μk = 0.2

Solution:
Given,
Weight of the block (force due to gravity) = 200 N
Force required to start the motion (static friction) = 80 N
Force required to keep the block moving at a constant velocity (kinetic friction) = 40 N
We can calculate the coefficients of friction as follows:
Coefficient of static friction (μs):
μs = (Force of static friction) / (Normal force)
μs = 80 N / 200 N
μs = 0.4
Coefficient of kinetic friction (μk):
μk = (Force of kinetic friction) / (Normal force)
μk = 40 N / 200 N
μk = 0.2
So, the coefficients of friction for the block and the surface are μs = 0.4 and μk = 0.2, respectively.

Question 14: A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pushing the box with a horizontal force of 35N. What is the magnitude of the force of static friction between the box and floor?
Answer: Friction force acting on box=35N

Solution:

Applied force =35 N Which is less than limiting friction.
∴Friction force acting on box=35N
Question 15: A man pulls a 50N box up a 30°incline to rest at a height of 4m. He exerts a total of 270J of work. What is the coefficient of friction on the incline?
Answer: 0.20

Solution:
Given,
The total work done = 270J
The force (F) = 50N
The angle (θ) = 30 degrees
The height (h) = 4m.
Coefficient of friction (μ)=?
Total work done = Work against gravity (W) + Work against friction(w)
Calculate the work against gravity(W):
W = 50sin30°*(4/sin30°)
W = 50N *1/2 * 8
W = 25*8
W = 200J
Calculate the work against friction(w):w
w= Total work done - Work done against gravity
w = 270J - 200 J
w = 70
Calculation of force of friction
F1=μN=μ50cos30°=μ50*(1.732/2)
Work against friction(w)=F1*(4/sin30°)
=F1*8
=μ50*(1.732/2)*8
=346.4*μ
∴346.4*μ =70
μ =70/346.4
μ = 0.20
So, the coefficient of friction on the incline is approximately 0.20.

Question 16: A 48kg crate is loaded onto a pick-up truck, and the truck speeds away without the crate sliding. If the coefficient of static friction between the truck and the crate is 0.4, what is the maximum acceleration that the truck can undergo without the crate slipping?
Answer: 3.92 m/s2

Solution:
Maximum static friction acting between the crate and the truck =F1
F1=0.4*48*9.8=188.16 N
Net force acting on the crate due to motion of trruck=F2
F2=48Xa (where a=acceleration of truck)
Crate is not moving so F1 and F2 are balanced forces.
48Xa=188.16
a=188.16/48 =3.92 m/s2

Question 17: : A 40kg box is initially sitting at rest on a horizontal floor with a coefficient of static friction μs=0.4. A horizontal pushing force is applied to the box. What is the maximum pushing force that can be applied without moving the box?
Answer: 156.96N

Solution:
Maximum static friction force that can be exerted on an object=0.4*40*9.8=156.96N
Therefore, the maximum pushing force that can be applied without moving the box is approximately 156.96 Newtons.

Question 18: A 5kg box slides across the floor with an initial velocity of 5m/s. If the coefficient of kinetic friction between the box and the floor is 0.1, how much time will it take for the box to come to a stop?
Answer:5.1 seconds

Solution:
Given,
m=5kg
μ=0.1
u=5m/s
v=0
Let time required to stop the box=t
The normal force acting on box=N=mg
N= 5 kg × 9.81 m/s^2 ≈ 49.05 N
Force of Kinetic Friction of box =μN= 0.1 × 49.05 N ≈ 4.905 N
This force will oppose the motion of object.
accelration =4.905/5=0.981 m/s^2 (This will be negative)
Equation of motion
a=(v-u)/t
-0.981=-5/t
t=5/0.981=5.1 second
So, it will take approximately 5.1 seconds for the box to come to a stop.

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