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Q1. When the block is displaced by a distance x, spring 1 exerts a force of k1x and spring 2 exerts a force of k2x. So, we have ma=k1x+k2x => ma=x(k1+k2)
Thus, a = x(k1+k2)/m. The direction will obviously be opposite to that of the displacement of the block as the springs compress/expand opposite to the direction of the applied force.
Q2. The man is at a height H. The two friends pull him up. The force behind this is the tension in the strings created by the 2 friends.
When he is at a height h, freeze that instant. So, for vertical equilibrium, we have:
2Tcos =mg. (m is the weight of the man). (TSin components cancel). (Theta is the angle between the vertical and the string)
So, T= mg/2cos. Now, in the triangle, we get cos = h/ h2 + d2/4 or cos@ = 2h/ 4h2+d2.
Putting this value, we get T = (mg/4h) 4h2+d2.
As the man moves up, h decreases, so, as T is inversely proportional to h, T, ie, the force applied by the friends increases as well.
Cheers!
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Moderation Team
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