Question

A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if is initially moving at a speed of 40 cm/s. If the frictional coefficient between the table and the block is 0.1 , how far does the block move before coming to rest?
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Here given in the question :-

Mass m = 250 g = 0.25 kg
Initial velocity , u = 40 cm/s = 0.4 m/sec.

Initial Kinetic Energy = 1/2 (mu²) 
= 1/2 (0.25 × 0.4²)
= 0.02 J.

Final Kinetic Energy = 0
Therefore the change in Kinetic energy = Work done by the frictional force
i.e = 0.02 J.

Now , frictional force = μmg
= 0.1 × 0.25 × 9.8 N
= 0.245 N.
assume that the block moves distance d before it comes to rest, Then
F × d = 0.02
d = 0.02/ 0.245 
d = 0.082 m
or d = 8.20 cm.


Hope it Helps.

Answer 2

Answer:

8.20cm!

Explanation:Here given in the question :-

Mass m = 250 g = 0.25 kg

Initial velocity , u = 40 cm/s = 0.4 m/sec.

Initial Kinetic Energy = 1/2 (mu²)  

= 1/2 (0.25 × 0.4²)

= 0.02 J.

Final Kinetic Energy = 0

Therefore the change in Kinetic energy = Work done by the frictional force

i.e = 0.02 J.

Now , frictional force = μmg

= 0.1 × 0.25 × 9.8 N

= 0.245 N.

assume that the block moves distance d before it comes to rest, Then

F × d = 0.02

d = 0.02/ 0.245  

d = 0.082 m

or d = 8.20 cm.

HOPE IT HEPL Y'ALL

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