In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of m, for which M=10 kg will move down with an acceleration of 2 m/s2, is: (take g=10 m/s2 and 37 =3/4) [figure]
- 4.5 kg
- 6.5 kg
- 9 kg
- 2.25 kg
Answer
(1) 4.5 kg
Solution
37 = 34 37 = 53 =0.6, 53 = 37 =0.8
M slides down and m slides up, friction on each block acts opposite to its motion.
Block M: Mg 53 - Mg 53 -T=Ma
80-15-T=20 T=45 N
Block m: T-mg 37 - mg 37 =ma
45=m(6+2+2)=10m m=4.5 kg
Answer: (1)