the acceleration of man of mass 40kg plus block of mass 10kg system, if man applies a force 300N on the string if the system is in horizontal and surfaces are smooth....
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Answers
Answer:
Thats ur answer
Explanation:
As per constant relation if acceleration of block B is
′
a
′
then
acceleration of block A is
′
2a
′
Now as per force equation of B
m
b
sin45−2T=m
b
a
T−m
a
gsin45=m
a
×2a
now solving above two equations for acceleration of the blocks
m
b
sin45−2m
a
sin45=(m
b
+4m
a
)×a
a=
m
b
+4m
a
(m
b
sin45−2m
a
sin45)g
now plug in all values
a=
40+40
(20
2
−10
2
)
a=6.6m/s
2
so acceleration of block B is 6.6m/s
2
and acceleration of block A is 13.2m/s
2
Given that,
Mass of man m₁= 40 kg
Mass of block m₂= 10 kg
Force = 30 N
We need to calculate the acceleration
Using formula of force
F=MaF=Ma
a=\dfrac{F}{M}a=
M
F
Where, M = mass of system
F = force
Put the value into the formula
a=\dfrac{30}{50}a=
50
30
a=0.6\ m/s^2a=0.6 m/s
2
Hence, The acceleration is 0.6 m/s²