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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
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Respuesta :

Answer:

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