help A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the s

Question

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A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the surface of a table, at the top of a theta = 40.0 degree 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 What is the velocity of the block as it leaves the incline? m/s How far from the table will the block hit the floor? m How much time has elapsed between when the block is released and when it hits the floor? S Does the mass of the block affect any of the above calculations? Yes No .

Explanation / 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

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