Question

FP 1. A sliding block of mass m 0.25 kg is subject to a force of magnitude 4 N that makes an angle of ф-30 with the horizontal surface. If the coefficient of kinetic friction between block and surface is 0.5, what is the resulting acceleration of the block along the surface Figure 1: Block on incline. 2. A block of mass m - 5 kg is subject to a force of magnitude 20 N that makes an angle of φ = 20 with the (frictionless) inclined plane of contact. The plane itself is inclined に30 with the horizontal (a) what is the acceleration ofthe block along the incline? (b) If the block begins at rest, in what direction will it movesul sequently i.e., up or down the incline)? (c) If there were friction with coefficient p,-0.2, then if initially at rest? would the block still slide
Questions 1&2

Answer 1

So the force is greater along downward direction so the body would more downward and downward acceleration is given as (25-18.79/5) = 1.24 m/s^2 (e) now it there exist frictional force it will act opposite to the moving direction frictional force [mg cos 90 degree - F sin 20 degree] Mu = [5 times 10 times squareroot 3/2 - 20 times 0.34] times 0.2 = 365 times 0.2 = 7.30 if the body moves downward required force will be (18.79 + 7.30)N = 26 but downward dragging force is 25N < 26N again if the body moves upward dragging force = 18.79 we have to over come (25 + 7.30)N = 32.30N so 18.79N < 32.30N as a result the body will remain still