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

Question

A block of mass m = 2.00 kg is released from rest at h = 0.800 m above the surface of a table, at the top of a ? = 25.0 A block of mass m = 2.00 kg is released from rest at h = 0.800 m above the surface of a table, at the top of a ? = 25.0 degree incline as shown in the figure below. The frictionless incline is fixed on a table of height H = 2.00 m. Determine the acceleration of the block as it slides down the incline. What is the velocity of the block as it leaves the incline? How far from the table will the block hit the floor? What time interval elapses between when the block is released and when it hits the floor? Does the mass of the block affect any of the above calculations?

Explanation / Answer

a) a =gsin25 = 4.14 m/s2

b)using energy conservation,

mv^2/2 = mgh

v = sqrt(2gh) = 3.96 m/s

c) in vertical :

u_y = 3.96sin25

H = u_yt +at^2/2

2 = 3.96sin25 x t + 9.81 t^2/2

4.9t^2 +1.67t - 2 = 0

solving this quadratic eqtn,

t = 0.49 sec

in horizontal :

d = u_xt = 3.96cos25 x 0.49 = 1.76 m

d) time taken on wedge -

v = u +at

3.96 = 0 + gsin25 x t

t = 0.956 sec

total time =0.956 + 0.49 = 1.45 sec

e) No.

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