As shown in the figure below, a light string that does notstretch changes from h

Question

As shown in the figure below, a light string that does notstretch changes from horizontal to vertical as it passes over theedge of a table. The string connects a 3.50 kg block, originally atrest on the horizontal table, 1.20 m above the floor, to a hanging2.10 kg block, originally 0.900 m above the floor. Neither thesurface of the table nor its edge exerts a force of kineticfriction. The blocks start to move with negligible speed. Considerthe two blocks plus the Earth as the system. Find the speed at which the sliding block leaves the edge ofthe table. Find the impact speed of the sliding block. How long must the string be if it does not go taut whilethe sliding block is in flight? Even with negligible kinetic friction, the coefficientof static friction between the heavier block and the table is0.540. Evaluate the force of friction acting on this block beforethe motion begins.

Explanation / Answer

Let small mass is m and heavier is M Smaller mass is .9 m above the ground. It has Potential Energy= mgh =2.1* 9.8*.9 This energy will convert into Kinetic energy of Heavier masswhen i t leave the edge So K.E. of heavier mass = P.E of smaller mass 1/2 Mv2 =mgh solve for v which comes out to be 3.25m/s for second part, the above velocity is initial vel for heavymass in downward direction. Let it final velocity isV(Impact) V2 -v2 = 2ad a= 9.8m/s2 d =1.2m v=3.25m/s V comes out to be 5.84m/s For next part from diagram I can say that string must be 1.2 mlong. Force of friction is smg=.50*3.5*9.8=18.52 newton d =1.2m v=3.25m/s V comes out to be 5.84m/s For next part from diagram I can say that string must be 1.2 mlong. Force of friction is smg=.50*3.5*9.8=18.52 newton

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