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

Question

A block of mass m = 2.00 kg is released from rest at h = 0.400 m from the surface of a table, at the top of a = 20.0° 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?

YesNo

Explanation / Answer

a) the acceleration down a frictionless incline is g sin(theta) = 9.8m/s/s x sin20 = 3.35m/s

b) we can find final velocity from

vf^2=v0^2 + 2ad

vf = final vel
v0=initial vel =0
a = accel = 3.35m/s/s
d= distance traveled = distance down the plane = 0.4m/sin20 = 1.16m
vf^2=0+2*3.35*1.16 => vf = 2.87m/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 = 2.87 cos 20 t
y(t) = 2 - v0 sin(theta) t - 1/2 gt^2 = 2- 2.87 sin 20 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 - 1.16 t - 4.9t^2
t=0.53s

the horizontal distance traveled in that time is

x = 2.87 cos 20 * 0.53s = 1.42m

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* 1.16 m/s / 3.35m/s/s] = 0.83s

so the total time = 0.83s+0.53s = 1.36s

No   

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