Question

EC1. A block of mass m-4.0 kg is put on top of a block of mass ms = 5.0 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table. a. What is the maximum possible acceleration of the 2 blocks together? (5) F word ood Ans. that can be applied to the lower block causing b. What is the maximum horizontal force this acceleration? (5) (1) ottobuli notioned maiting muisdienser. 8 door 'blood lb Ans.

Answer 1

m m Given, Mass of the top block, m = 4.0kg Mass of the bottom lock, my = 5.0 kg The force act on the top block , F = 12N Free body diagram of top block is given by mi → 12N fe is the maximum static frictional force that privant the block from slipping The maximum force acts on the top block to accelerated is 12N. using the 2nd Newtons law on top block in x dinletion. {F = maa 12N = 4kg. ax ax = 12 jan= 3 m/s² The acceleration of the bottom block. e same as the acceleration of the to block because they accelerated together as a system.

The free body diagram of a bottom block a FN law in x dire using Newton's and E Fx = max F-12=5*3 F=15+12 FE2ZN . (as Maximum possible acceleration of the 2 blocks together is 3.0 m/s2 (6) Maximum horizontal force, ř that can be applied to the lower block casusing this acceleration is 27N