Questions
A block A of mass rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass is suspended. The coefficient of kinetic friction between the block and the table is . When the block A is sliding on the table, the tension in the string is
detailed solution
Correct option is A
Given situation is shown in the figure. Here, N=m1g f=μkN=μkm1g . . . . .(i)Let α be the acceleration of blocks. Equation of motion for A and BT-f=m1a . . . . .(i) m2g-T=m2a . . . . .(ii)Adding equation(ii) and (iii), we get.m2g-f=m1+m2aa=m2g-fm1+m2Put this value of α in equation (iii) T=m2g-m2m2g-fm1+m2=m1m2g+m1m2μkgm1+m2=m1m21+μkgm1+m2 (using . (i))Talk to our academic expert!
Similar Questions
A constant force F = m2g/2 is applied on the block of mass m1 as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
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