a) A very slippery block of ice slides down a smooth ramp tilted at angle pheta.The ice is released from rest at vertical height h above the bottom of the ramp. Find an expression for the speed of...

a) A very slippery block of ice slides down a smooth ramp tilted at angle pheta.

The ice is released from rest at vertical height h above the bottom of the ramp. Find an expression for the speed of the ice at the bottom.

b) Evaluate your answer to part a) for ice released at a height of 30cm on ramps tilted at an angle of 20 degrees and at 40 degrees.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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When the block of ice is at the op of the ramp due to the slope of the ramp component of its weight act parallel to the ramp. This cause the ice block to move along the ramp and come to the bottom of the ramp.

By using F= ma along the ramp;

Let us consider the mass of ice block as m kg.

mg`sintheta` = m*a

          a = g`sintheta`

 

Using `v^2 = u^2-2as` along the ramp for the ice cube;

s = distance that the ice cube travel along ramp

sin`theta` = h/s

          s = h/sin`theta`

length measured upwards taken as positive. So here length as measured top to bottom or downward.

     So s = -h/sin`theta`

      v^2 = 0-2*g`sintheta` *(-h/`sintheta` )

          v = `sqrt(2gh)`

 

So here we notice that the final velocity is not depending on the angle of the ramp.

When h=30cm

     v = `sqrt(2gh)`

        = `sqrt(2*9.81*30/100)`

        = 2.426 m/s

 

So since there is no effect from the angle the final velocity of the ice block at bottom of the ramp is 2.426 m/s.

 

There is another easy method to do this. It is by using energy.

At the top of the ramp the ice bock will have potential energy. This potential energy convert to kinematic energy at the bottom.

If the datum is bottom of the ramp;

Initial potential energy of  ice block = m*g*h

final kinematic energy of ice block   = `1/2*m*v^2`

 

Since there is no energy losses;

Initial potential energy = final kinematic energy

                        mgh  = `1/2mv^2`

                              v = `sqrt(2gh)`

 

We can see we get the same answer as in the first method.

 

Assumption;

  • There is no frictional forces acting on the ice block since it is very slippery
  • While moving the mass of ice remain constant

 

 

Sources:

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