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a chord is used to lower vertically a block of mass M through a distance d at a constant downward accn g/3 d work done by d chord on d block is ?
a ballon is going vertically upwards with a velocity of 10m/s 75m above d ground a stone is released d time taken by d stone to reach d ground is ?
a body falling freely covers 9/25 of a dist in its last sec d height of d point from d ground is ?
a ballon rises from rest with const accn g/8 a stone is released from it when it has risen to a height h. d time taken by d stone to reach d ground is ?
a heavy steel ball of mass >1kg moving with a velocity of 2m/s collides head on with a stationary ping-pong ball of mass less dan 1gm .d collision is elastic after collision d ping-pong ball moves approximately with a velocity of
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5 Answers
3F.S=workdone
then the force due to tension does work in lowering the block.
T.S=net work find tension.
106Q1. a = g/3 downward.
mg - T = ma
So, T = 2mg/3
tension acts opposite to block's displacement
So, work done by tension = T.S =TScosÎ
= 2Mg/3 * d * (-1)
= -2Mgd/3
Q2. When the stone is released it has the same velocity as the balloon but the acceleration becomes g downward. Taking downward direction as positive:
u get
75 = -10t + 5t2 (h=-ut+at2/2)
Solving it you will get t=5 sec. other solution is inadmissible as get time <0
Q4. as i said when the stone is released it acquires the velocity of the balloon and not the acceleration
V of balloon at height h = V2 = 2*g/8*h = gh/4
So, V = √gh/2
Now using the method of sign convention as above
u will get
h = -√ght/2 + gt2/2
u can solve and get t