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 theta = 20.0 degree incline as shown in the figure 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/s^2 (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? Yes No

Explanation / Answer

(a) The acceleration down a frictionless incline is

g sin(theta) = 9.8m/s*sin(20 deg) = 3.35 m/s^2

(b) Let length of incline, L

sin(theta) = h/L

L = h/sin(theta)

L = 0.4/sin(20 deg) = 1.169 m

Initial velocity, u = 0

Displacement, L = 1.169 m

Acceleration a = 3.35 m/s^2

Final velocity v = ?

v^2 = u^2 + 2aL

v^2 = 0 + 2*3.35*1.169 = 7.83

v = sqrt(7.83) = 2.79 m/s

(c) Consider the motion from the bottom of the incline to the floor.
Take downward direction as positive.

In vertical direction

Initial velocity = v sin(theta)

Acceleration = g

Height = H

Let time taken = t

H = v sin(theta) * t + 1/2 gt^2

2 = 2.79* sin(20 deg) * t + 1/2 * 9.8 * t^2

2 = 0.954 t + 4.9 t^2

4.9 t^2 + 0.954 t - 2 = 0

t = [-0.954 +- sqrt(0.954^2 + 4*4.9*2)]/(2*4.9)

t cannot be negative.

Therefore

t = 0.55 s

Horizontal component of velocity = v cos(theta)

This is constant because there is no force in horizontal direction.

R = v cos(theta) * t

R = 2.79* cos(20 deg) * 0.55

R = 1.44 m

(d) Let t1 = time taken by the block to move to the bottom of the incline.

Distance covered, L = 1.169 m

Initial velocity u = 0

Final velocity, v = 2.79 m/s

Average velocity, Vavg = (v+u)/2

Vavg = (2.79+0)/2 = 1.395 m/s

t1 = L/Vavg

t1 = 1.169/1.395 = 0.838 s

Time taken to move from bottom of the incline to the floor = 0.55 s

Therefore, total tie taken from the time the block is released to the time it hits the floor = 0.838 + 0.55 = 1.388 s

(e) The mass of the block has not been used in making any of the above calculations. Therefore, it does not affect.

No

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