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19.A man sits on a chair supported by a rope passingover a frictionless fixed pulley. The man who weighs1000 N exerts a force of 450 N on the chairdownwards, while pulling the rope on the other side.If the chair weighs 250 N, then the acceleration of thechair is(1) 0.45 m/s^2(3) 2 m/s^2(2) zero(4) 9/25m/s^2 |
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Answer» Suppose acceleration is in direction of man going downwards so by drawing the diagram and using fbd consider g to be 10 so the mass of the person is 100kg and 125kg of the chair , we get the following 2 equations 1 T- 450N (force applied by the man) = 45a (450÷g i.e mg÷mg = m) and 2- [mg (of man) + mg of chair = (mass of man + mass of chair) so putting values the equation are as follow 1)T - 450N = 45a , 2) 1250 - T = 125 a. Solving equation via elimination method we get a= 4.70 |
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