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Find velocity of block 'B' at the instant shown in figure.
                  
  • a)
    25 m/s 
  • b)
    20 m/s 
  • c)
     22 m/s
  • d)
    30 m/s
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Find velocity of block 'B' at the instant shown in figure. ...
Net Force = Force exerted by Cyclist - Frictional Force
Also, according to newton's second law
Fnet = m.a
250 N - Frictional Force = 30x4
∴ Frictional Force = 250 - 120 N
= 130 N
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Most Upvoted Answer
Find velocity of block 'B' at the instant shown in figure. ...
Net Force = Force exerted by Cyclist - Frictional Force
Also, according to newton's second law
Fnet = m.a
250 N - Frictional Force = 30x4
∴ Frictional Force = 250 - 120 N
                             = 130 N
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Community Answer
Find velocity of block 'B' at the instant shown in figure. ...
There are two methods to solve these type questions easily. One method is that, velocity along a string remains constant and the other way to do is that the sum of product tensions and velocity is zero (taking sign into consideration). If velocity and tension are in same direction positive sign and if there are in opposite directions negative sign is taken.
Note that in the first method velocity of the string wound around P is taken 20 m/s because if block A moves one unit downward then pulley P moves two units downward so if A's velocity is 10 m/s then P's velocity will be 20 m/s. 
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Find velocity of block 'B' at the instant shown in figure. a)25 m/sb)20 m/sc)22 m/sd)30 m/sCorrect answer is option 'A'. Can you explain this answer?
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