he acceleration of M is (g10 ms 2)F-50N30-sinIM-10kg320 2-mS17 232一m

1.	he acceleration of M is (g10 ms 2)F-50N30-sinIM-10kg320 2-mS17 232一ms3(d) c(b) ms
Answer» the force acting in horizontal direction is = 50cos∅ and ∅ = 37° so, Fh = 50(4/5) = 40N and the vertical component of force will try to reduce the normal reaction so, Normal reaction will be = mg -Fv = 100-(50sin37) = 100-(503/5) = 70N now frictional force is μN = 1/3(70) = 70/3 N so, equation along x direction is Net force = massa => a*(10) = 40-70/3 = 50/3 => a =(5/3) m/s². option D should be the answer.

he acceleration of M is (g10 ms 2)F-50N30-sinIM-10kg320 2-mS17 232一ms3(d) c(b) ms

Answer»

the force acting in horizontal direction is = 50cos∅

and ∅ = 37°

so, Fh = 50*(4/5) = 40N

and the vertical component of force will try to reduce the normal reaction

so, Normal reaction will be = mg -Fv = 100-(50sin37) = 100-(50*3/5) = 70N

now frictional force is μN = 1/3*(70) = 70/3 N

so, equation along x direction is

Net force = mass*a

=> a*(10) = 40-70/3 = 50/3 => a =(5/3) m/s².

option D should be the answer.