ffigure shows a uniform rod of mass 3 kg and of length 30 cm is tension figure 8 by constant forces of 20 Newton 32 Newtons the acceleration of the road is
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4m/s^2 is the answer??
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Answered by
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according to Newton's law of motion,
body moves where force exists . here in right side force { 32N } is greater than left side side {20N}. so, body of mass 3kg moves in rightward.
Let the force exerted by 20cm part of the rod on the 10cm part is N.
first we have to find mass of 20cm part and 10cm part of body .
mass of 20cm part , M = 3kg × 20cm/(10cm + 20cm) = 2kg
mass of 10cm part , m = 3kg - 2kg = 1kg
now, use Newton's 2nd law ,
for 20cm part ,
forward force - backward force = Ma
32 - N = 2a -------(i) [ here a is acceleration]
similarly for 10cm part,
N - 20 = a -------(ii)
from eqs. (i) and (ii),
12 = 3a => a = 4 m/s²
hence, acceleration of rod is 4 m/s²
body moves where force exists . here in right side force { 32N } is greater than left side side {20N}. so, body of mass 3kg moves in rightward.
Let the force exerted by 20cm part of the rod on the 10cm part is N.
first we have to find mass of 20cm part and 10cm part of body .
mass of 20cm part , M = 3kg × 20cm/(10cm + 20cm) = 2kg
mass of 10cm part , m = 3kg - 2kg = 1kg
now, use Newton's 2nd law ,
for 20cm part ,
forward force - backward force = Ma
32 - N = 2a -------(i) [ here a is acceleration]
similarly for 10cm part,
N - 20 = a -------(ii)
from eqs. (i) and (ii),
12 = 3a => a = 4 m/s²
hence, acceleration of rod is 4 m/s²
Answered by
4
Answer:4m/s^2
Explanation:
M1=3kg×20cm/10+20
M1=60/30
M1=2kg
M2=3kg-2kg
M2=1kg
You make a equation of constant force
32-N=2a
20-N=a
Substring eqEction
12=3a
Answer is 4m/s^2
This question is motion module
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