Solving Numerical Problems involving Frictional Force or friction – for class 11

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Set 2

Problem iii)
A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity.
(a) What is the coefficient of sliding friction between the block and the tabletop?
(b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity?

Solution

a) Weight of the block = W = mg = 40N
In this case the normal force = FN = W = 40N
Let the coefficient of sliding friction between the block and the tabletop = μ

Then, sliding Friction between the block and the tabletop = FF= μFN = μ 40 = 40μ
As per the question, a force of 14 N is just required to keep the block moving at a constant speed.

Hence, sliding friction magnitude = applied force magnitude
40μ = 14
μ = 14/40 = 0.35

b) In the second case, the normal force = FN = W = (40 + 20) N= 60 N
Normal force = FN = W = 60N

Friction = FF= μFN = 0.35 x 60 =21 N
Hence this time to move at a constant velocity the applied force has to be equal to 21 N

Solving Numerical Problems involving Frictional Force or friction – for class 11
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