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

Answer 1

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