A 2.00 - kg - box, starting from rest, slides down a friction- less incline. The

Question

A 2.00 - kg - box, starting from rest, slides down a friction- less incline. The incline makes and angle of ? = 30 ? from the horizontal and is fixed to a table. The incline has a maximum height of h = 0.500 m above the table that is itself H = 2.00 m above the ground. (a) Find the accleration of the box during the sliding period. (b) What is the velocity of the box as it leaves the table. (c) What horizontal distance (R) from the table will the block reach when it hits the ground.  (d) How much time elapses between when the box is released and when it hits the ground? (e) Does any of the above calcula- tions depend on the mass of the box? A 2.00 - kg - box, starting from rest, slides down a friction- less incline. The incline makes and angle of ? = 30 ? from the horizontal and is fixed to a table. The incline has a maximum height of h = 0.500 m above the table that is itself H = 2.00 m above the ground. (a) Find the accleration of the box during the sliding period. (b) What is the velocity of the box as it leaves the table. (c) What horizontal distance (R) from the table will the block reach when it hits the ground.  (d) How much time elapses between when the box is released and when it hits the ground? (e) Does any of the above calcula- tions depend on the mass of the box?

Explanation / Answer

Acceelration during Sliding = gSin(theta)

= 9.8*Sin(30 degree)

= 4.9 m/sec^2

Here

Loss in potential Energy = Gain in Kinetic Energy

0.5*m*v^2 = mgh

v sqrt(2gh)

= sqrt(2*9.8*0.5)

= 3.13 m/sec

Here

Horizontal Component of Velocity = 3.13*Cos(30 degree)

= 2.71 m/sec

Vertical Component = 3.13*Sin(30)

= 1.565 m/sec

Time taken to hit the ground let be t

Therefore

2 = 1.565*t + 0.5*9.8*t^2

t = 0.4988 sec

Therefore

R = Horizontal Component*t

= 0.4988*2.71

= 1.35 m

No it does not depend on Mass

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