18 Laws of Motion
8.
Calculate acceleration of mass 2 kg if a constant
force of 12 N is applied on 6 kg mass given that the
surface between the blocks is rough with u = 0.3
while ground surface is smooth.
13.
u=0.3
2kg
r
6kg
► F=12 N
r
(1) 1.5 m/s2
(2) 3 m/s2
(4) 1 m/s2
(3) Zero
14
Answers
Answer:
The acceleration of the block is 1 m/s²
Therefore, option (1) is correct.
Explanation:
Given:
Force F_1=2\sqrt{2}F
1
=2
2
N acting on block of mass m=3m=3 kg
acting at an angle \theta=45^\circθ=45
∘
Another force F_2=1F
2
=1 N is acting on the block
Resolving the force F_1F
1
in horizontal and vertical direction
The horizontal component
=F_1\cos\theta=F
1
cosθ
=2\sqrt{2}\cos 45^\circ=2
2
cos45
∘
=2\sqrt{2}\times\frac{1}{\sqrt{2}}=2
2
×
2
1
=2=2 N
The vertical component
=F_1\sin\theta=F
1
sinθ
=2\sqrt{2}\sin 45^\circ=2
2
sin45
∘
=2\sqrt{2}\times\frac{1}{\sqrt{2}}=2
2
×
2
1
=2=2 N
The vertical force will be balanced by the weight of the body and the nrmal reaction
The net horizontal force will be
F=2+1=3F=2+1=3 N
From the second law of Newton
We know that
F=maF=ma
Where mm is mass and aa is acceleration
Therefore,
a=\frac{F}{m}a=
m
F
\implies a=\frac{3}{3}=1⟹a=
3
3
=1 m/s²
Hope this answer is helpful.
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