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Question

A block of mass 2 kg placed on a rough horizontal surface having coefficient of friction 0.25 is acted upon by a force of
10 N as shown below. The distance travelled by the block in the first five seconds is


A
16.75 m
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B
31.25 m
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C
62.5 m
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D
Zero
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Solution

The correct option is B 31.25 m

Mass, m=2kg
External force, F=10N
Normal reaction, N=20N
Frictional force, f=μR
Here, μ = coefficient of friction = 0.25
R=mg=2×10=20N
f=μR=0.25×20=5N
Net force acting on the block, Fnet=Ff
= 10 N – 5N
= 5 N
According to second law of motion:
Fnet=ma=5N
a=5m=52=2.5m/s2

Since initially the block is not moving, its initial velocity, u is '0'.
Distance travelled by the block in 5sec,
s=ut+12at2
s=0×t+12(52)(5)2
s=1254=31.25m
The displacement in the first five seconds is 31.25 m.
Hence, the correct answer is option (2).

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