Question #4338

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 θ = 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.
(b) What is the velocity of the block as it leaves the incline?
(c) How far from the table will the block hit the floor?
(d) How much time has elapsed between when the block is released and when it hits the floor?

Expert's answer

If the inclined plane has an incline of 30 degrees and is 1/2m high, then we have tan(30) = .5/hyp, hyp = .5/tan(30), hypotenuse = 0.866m. That's the length of the plane.

a) F = ma, so a = F/m = 9.8*cos(60)/2kg = 4.9m/s^2

Use the pythagorean theorem and you get the 3rd side of the inclined plane's triangle is 1m

b) dist = 1/2at^2, sqrt(2a/m) = t, sqrt(2(4.9)/2) = t = sqrt(4.9) = 2.21sec

v = at = 4.9(1) = 4.9m/s

c.) horizontal component of speed is 4.9 * cos(30) = 4.24m/s

vertical component of speed is 4.9 * sin(30) = .5m/s

d = 1/2at^2, where d = height of table = 2.0m

a = 9.8m/s^2, so

t = sqrt(2(2)/9.8) = 0.639s until it hits the floor

d = vt = 4.24*0.639 = 2.7m from the table

d) 2.7 + 2.21 = 4.9 sec.

e) No, mass does not affect how quickly something falls, and everything here is frictionless.

a) F = ma, so a = F/m = 9.8*cos(60)/2kg = 4.9m/s^2

Use the pythagorean theorem and you get the 3rd side of the inclined plane's triangle is 1m

b) dist = 1/2at^2, sqrt(2a/m) = t, sqrt(2(4.9)/2) = t = sqrt(4.9) = 2.21sec

v = at = 4.9(1) = 4.9m/s

c.) horizontal component of speed is 4.9 * cos(30) = 4.24m/s

vertical component of speed is 4.9 * sin(30) = .5m/s

d = 1/2at^2, where d = height of table = 2.0m

a = 9.8m/s^2, so

t = sqrt(2(2)/9.8) = 0.639s until it hits the floor

d = vt = 4.24*0.639 = 2.7m from the table

d) 2.7 + 2.21 = 4.9 sec.

e) No, mass does not affect how quickly something falls, and everything here is frictionless.

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