Any block of mass 2kg is placed on the horizontal road. The coefficient of friction between the block and the road is 0.2, if 6n horizontal force is applied on the block, find the distance traveled in the first second
Answers
given that ,
》mass of block = 2kg
》force applied = 2.5
》coefficient of static friction, µs = 0.4
☆Maximum static frictional force,
fmax = µs × m × g = 0.4 × 2 × 9.8
=> fmax = 7.84 N
Since,
the applied force is less than the maximum static frictional force,
the frictional force on the block is equal to the applied force = 2.5 N.
☆it is due to the fact that static friction is a self adjusting force
Explanation:
The various forces acting on the block is as shown in the figure.
Here, m=2 kg,μ=0.1,F=6 N,g=10 ms
−2
Force of friction,
f=μN=0.1×2 kg×10 ms
−2
=2 N
Net force with which the block moves
F
′
=F−f=6N−2N=4N
Net acceleration with which the block moves
a=
m
F
′
=
2kg
4N
=2 ms
−2
Distance travelled by the block in 10 s is
d=
2
1
at
2
=
2
1
×2 ms
−2
(10)
2
=100 m(∴u=0)
As the applied force and displacement are in the same direction therefore angle between the applied force and the displacement is θ=0
∘
.
Hence, work done by the applied force,
W
F
=Fdcosθ=(6N)(100m)cos0
∘
=600 J.
solution
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