As shown in the sketchy below, a block of mass M = 2.0kg, which was initially at

Question

As shown in the sketchy below, a block of mass M = 2.0kg, which was initially at rest (vi = 0) slides down a frictionless curved ramp. The top of the ramp is 2.0 meters above the ground, and the bottom of the ramp is 0.2 meter above the ground. The block leaves the ramp horizontally at speed v and lands on the ground a horizontal distance d away. Approximate g, gravity, by 10 m/s^2.

i) Solve for the block's kinetic energy K and its speed v when it leaves the ramp.

ii) Solve for the distance d and for the block's final kinetic energy (Kf) when it hits the ground.

Explanation / Answer

i) let, Mass of the block M = 2.0kg

    Distance from top of the ramp to ground be h1 = 2.0 m

    Distance from bottom of the ramp to ground be h2 = 0.2 m

   Using law of conservation of energy, As block slides from top of ramp to bottom of ramp

    Potential energy = Kinetic energy

     Mg(h1 - h2) = (1/2)Mv2

       g(h1 - h2) = (1/2)v2

       10(2 - 0.2) = (1/2)v2

         2(18) = v2

         v = 6.0 m/s (This the velocity of block , when it leaves the ramp)

        Also , Kinectic energy = Mg(h1 - h2)

                                             = 2.0(10)(2 - 0.2)

                                             = 36 J

ii) As block leaves the ramp, velocity of block is horizontal, vx = v = 6.0 m/s

and yth component of velocity is vy = 0 m/s

Using Kinematical equation,

s = ut + (1/2)at2    , we can find time to reach ground

h1 = vyt + (1/2)(-g)t2 (where, s = h1 , u = vy , a = -g)

-0.2 = (1/2)(-10)t2

    t2 = 4

    t = 2 sec

Horizontal Distance d = vxt

                                 d = (6.0)(2)

                                  d = 12 m

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