Q#1
A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s². What is the coefficient of kinetic friction between the block and the plane ?Answer:
The friction force causing the deceleration = f = m.a
Where m = mass of the body,
and a = deceleration of the body = 4.0 m/s²
So f = 4m
Normal force on the body = weight of the body = mg N
So coefficient of kinetic friction = 4m/mg = 4/g = 4/9.8 = 0.4
Q#2
A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest ?
Answer:
Let mass of the block be m . Weight of block = mg
So, Normal force on the block N = mg, Given µ = 0.10
Force of friction f = µN = (0.10)mg
This force of friction will try to stop the block with retardation
a = Force/mass
= (0.10)mg/m = 0.10g = 0.98 m/s²
Now initial velocity of the block = u = 10 m/s
Final velocity = 0, a = -0.98 m/s², s = distance traveled =?
Using the relation v² = u²+2as
we get,
0² =10² - 2(0.98)s
S = 100/1.96 m = 51 m
Q#3
A block of mass m is kept on a horizontal table. If the static friction coefficient is µ , find the frictional force acting on the block ?
Answer:
Since the block is only kept on the horizontal surface and no force is applied on it trying to move, so force of friction is zero because frictional force comes in to action only when the surfaces in contact are in relative motion or trying to slip over each other.
Q#4
A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in first two seconds. Find the coefficient of kinetic friction between the two.
Answer:
Let us first find the acceleration of the block = a
We have, initial velocity = u = 0, Distance traveled = s = 8 m
time = t = 2 s, From the relation s=ut+½at²,
we get
8 = 0 + ½a(2)² = 2a
a = 4 m/s²
So the net force on the block along the incline = ma = 4m
= mgsin30° – f
= ½mg – f
Equating these two,
½mg – f = 4.m
f = ½(9.8m) – 4m
= 4.9m – 4m = 0.9m
Normal force on the block
= mgcos30° = ½√3 x 9.8m = 8.48m
So coefficient of kinetic friction = 0.9m/8.48m = 0.11
Q#5
Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest? The mass of block is 4 kg.
Answer:
In the previous problem when there is no external force, net force along the incline is =mg.sin30°-f N where m = mass of the block. In this problem when m = 4 kg and and external force along the incline = 4 N, f = 0.9m,
Net force along the incline
= mgsin30° – f + 4
= 4(9.8)½ – 0.9(4) + 4 = 20 N
So acceleration along the incline, a = Force/mass = 20/4 = 5 m/s²
Given t = 2 s, using the relation, distance traveled
s = ut + ½at²
= 0 + ½(5)(2)² = 10 m
Q#6
A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.
Answer:
The only force acting on the block trying to move it down the plane is the component of its weight = mgsinθ
{where θ = 30° (angle of inclination) and m = 2 kg (mass of the block)}
= 2(9.8)sin30° = 9.8 N
To move the block in either direction the net force along the plane in that direction must be just more than the limiting force of friction So let us first find this Limiting frictional force f which is equal to µR where R = Normal force on the block.
Normal force on the block is equal to component of the weight perpendicular to the plane
= mgcos θ = 2(9.8)cos30° = 2(9.8)(½√3) = 9.8√3
Limiting frictional force f = µR = 0.2(9.8√3) = 3.394 N
(Given µ = 0.2)
(a) To move the block up the incline
Net force opposing the movement = Limiting frictional force + Component of weight along the incline
= 3.394 + 9.8 = 13.194 N
So to move the block up the incline should be just more than this force = 13.194 N ≈ 13.2 N
(b) To move the block down the incline
Since the component of weight which tries to move the block down the plane is more than the limiting frictional force (9.8 > 3.394), so no external force is needed to move the block down the incline. It will move itself.
i.e. external force needed to move the block down the incline = zero.
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