Questions
A mass of 3 kg descending vertically downwards supports a mass of 2 kg by means of a light string
passing over a pulley. At the end of 5 s, the string breaks . How much high from now the 2 kg mass
will go? (Take, 9 = 9.8 )
detailed solution
Correct option is A
Acceleration of system before breaking the string,a= Net pulling force Total mass =3g-2g5=g5After 5 s velocity of system, v=at=g5×5=g ms-1Now, h=v22g=g22g=g2=49 mTalk to our academic expert!
Similar Questions
A constant force F = m2g/2 is applied on the block of mass m1 as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
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