A wooden block of mass m = 1.8 kg starts from rest on an inclined plane sloped a

Question

A wooden block of mass m = 1.8 kg starts from rest on an inclined plane sloped at an angle q = 23o from the horizontal. The block is originally located 1.5 m from the bottom of the plane. Assume the inclined plane is frictionless.

1) What is the magnitude of the component of the block's weight parallel to the incline?

2)What is the magnitude of the component of the block's weight perpendicular to the plane?

3)What is the magnitude of the normal force on the block?

4)What is the net force on the block?

5)What is the acceleration of the block?

6)How long does it take the block to reach the bottom of the incline?

Explanation / Answer

a). mg sin(theta) = 1.8* 9.8 *sin23 = 6.892 N b). mg cos(theta) = 16.238 N c). mg cos(theta) = 16.238 N d). F = mg sin(theta) = 6.892 N e). g sin 23 =3.829 m/s2 f). s =u*t +0.5a*t^2 => 1.5 = 0+ 0.5 *3.829* t^2 => t = 0.885 s

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