19. In fig. m = 5kg. m = 2kg and F= IN. Find the acceleration of either block. Describe the motion of
if the string breaks but Fcontinues to act.
a) 4 m/s. 8 + 0.2 m/s
b) 4.3 m/s. m/s
c) 4.1 m/s, & + 0.2 m/s
d) 4.3 m/s. g = 0.2 m/s
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Answers
Answer:
m1 = 5 kg,
m2 = 2 kg and
F = 1 N
Let the acceleration of the blocks be a.
From the free-body diagram,
m1a = m1g + F − T …(i)
Again, from the free-body diagram,
m2a = T − m2g − F …(ii)
Adding equations (i) and (ii), we have:
a=g[ m1-m2/ m1+m2 ]
⇒a=3g/7=29.4/7
=4.2 m/s2
Hence, acceleration of the block is 4.2 m/s2.
After the string breaks, m1 moves downward with force F acting downward. Then,
m1a = F + m1g
5a = 1 + 5g
⇒a=5g+1/5
=g+0.2 m/s²
m1 = 5 kg,
m2 = 2 kg and
F = 1 N
Let the acceleration of the blocks be a.
From the free-body diagram,
m1a = m1g + F − T …(i)
Again, from the free-body diagram,
m2a = T − m2g − F …(ii)
Adding equations (i) and (ii), we have:
a=g[ m1-m2/ m1+m2 ]
⇒a=3g/7=29.4/7
=4.2 m/s2
Hence, acceleration of the block is 4.2 m/s2.
After the string breaks, m1 moves downward with force F acting downward. Then,
m1a = F + m1g
5a = 1 + 5g
⇒a=5g+1/5
=g+0.2 m/s²
m1 = 5 kg,
m2 = 2 kg and
F = 1 N
Let the acceleration of the blocks be a.
From the free-body diagram,
m1a = m1g + F − T …(i)
Again, from the free-body diagram,
m2a = T − m2g − F …(ii)
Adding equations (i) and (ii), we have:
a=g[ m1-m2/ m1+m2 ]
⇒a=3g/7=29.4/7
=4.2 m/s2
Hence, acceleration of the block is 4.2 m/s2.
After the string breaks, m1 moves downward with force F acting downward. Then,
m1a = F + m1g
5a = 1 + 5g
⇒a=5g+1/5
=g+0.2 m/s²
m1 = 5 kg,
m2 = 2 kg and
F = 1 N
Let the acceleration of the blocks be a.
From the free-body diagram,
m1a = m1g + F − T …(i)
Again, from the free-body diagram,
m2a = T − m2g − F …(ii)
Adding equations (i) and (ii), we have:
a=g[ m1-m2/ m1+m2 ]
⇒a=3g/7=29.4/7
=4.2 m/s2
Hence, acceleration of the block is 4.2 m/s2.
After the string breaks, m1 moves downward with force F acting downward. Then,
m1a = F + m1g
5a = 1 + 5g
⇒a=5g+1/5
=g+0.2 m/s²
m1 = 5 kg,
m2 = 2 kg and
F = 1 N
Let the acceleration of the blocks be a.
From the free-body diagram,
m1a = m1g + F − T …(i)
Again, from the free-body diagram,
m2a = T − m2g − F …(ii)
Adding equations (i) and (ii), we have:
a=g[ m1-m2/ m1+m2 ]
⇒a=3g/7=29.4/7
=4.2 m/s2
Hence, acceleration of the block is 4.2 m/s2.
After the string breaks, m1 moves downward with force F acting downward. Then,
m1a = F + m1g
5a = 1 + 5g
⇒a=5g+1/5
=g+0.2 m/s²