A 50g mass is placed on a straight air track sloping at an angle of 45° to the horizontal. Calculate in metre per second, the acceleration of the load as it slides down and also the distance it will be moved from rest in two seconds

a = 0.347 m/s^2 and S = 0.694 m

Explanation:

on a slope which is friction less acceleration is m g sin Ф whereФ is angle of inclined plane with horizontal

here m = 50g = 0.050 kg, g = 9.81 m/s^2, Ф = 45°

a = 0.050 * 9.81 * sin45°

a = 0.347 m/s^2

now

S = Ut + (1/2) at^2

where S is displacement U is initial velocity, a is acceleration and t is time

S = (0) (2) + (1/2) (0.347) (2) ^2

S = 0 + (0.5) (0.347) (4)

S = 0.694 m