A block of mass m = 2.00 kg is released from rest at h = 0.400 m above the surfa

Question

A block of mass m = 2.00 kg is released from rest at h = 0.400 m above the surface of a table, at the top of a ? = 40.0

A block of mass m = 2.00 kg is released from rest at h = 0.400 m above the surface of a table, at the top of a ? = 40.0 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) the acceleration down a frictionless incline is g sin(theta) = 9.8m/s/s x sin40 = 6.30m/s

b) we can find final velocity from

vf^2=v0^2 + 2ad

vf = final vel

v0=initial vel =0

a = accel = 6.30m/s^2

d= distance traveled = distance down the plane = 0.4m/sin40 = 0.622m

vf^2=0+2*6.30*0.622 => vf = 2.80m/s

now we write the x and y equations of motion for the block; down will be negative, so

x(t) = v0 cos(theta) t = 2.80 cos 40 t

y(t) = 2 - v0 sin(theta) t - 1/2 gt^2 = 2- 2.89 sin 40 t - 4.9 t^2

to find the time it takes the block to hit the ground once it leave the incline, set y(t)=0

and solve the quadratic 0=2 - 1.80 t - 4.9t^2

t=0.48s

the horizontal distance traveled in that time is

x = 2.80cos 40 * 0.48s = 1.029m

the time for the block to reach the bottom of the incline can be found from

d = 1/2 at^2 or t= sqrt[2d/a] = sqrt[2* 0.622 / 6.30] = 0.44s

so the total time = 0.44s+0.48s = .92s

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