A block of mass m = 2.00 kg is released from rest at h = 0.800 m from the surfac

Question

A block of mass m = 2.00 kg is released from rest at h = 0.800 m from the surface of a table, at the top of a = 30.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m.

(a) Determine the acceleration of the block as it slides down the incline.
m/s2

(b) What is the velocity of the block as it leaves the incline?
m/s

(c) How far from the table will the block hit the floor?
m

(d) What time interval elapses between when the block is released and when it hits the floor?
s

(e) Does the mass of the block affect any of the above calculations?

Choose:

1.Yes

2.No   

Explanation / Answer

a] acceleration a = g sin theta = 9.8* sin 30 degree = 4.9 m/s^2

b] length of incline = h/sin 30 degree = 0.8/0.5 = 1.6 m

by third equation of motion,

v = sqrt(2as) = sqrt(2*4.9*1.6) = 3.96 m/s

c] time taken to fall hit floor, H = ut + 0.5gt^2

2 = 3.96 sin 30 degree t + 0.5*9.8 t^2

t = 0.468 s

distance from floor = Vx*t = 3.96* cos 30 degree *0.468

= 1.6 m

d]total time = 3.96/4.9 + 0.468 = 1.276 s

e] no

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