Question

A very slippery block of ice slides down a smooth ramp tilted at angle ? The ice is released from rest at vertical height h above the bottom of the ramp.

A)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 22cm on ramp tilted at 40 ??

Answer 1

The speed of the ice at the bottom of a frictionless ramp can be found using conservation of energy with the formula \(v = \sqrt{2gh}\). For a height of 0.22 meters, the speed would be approximately 2.083 meters per second.

Step-by-step explanation:

A block of ice sliding down a smooth ramp will increase in speed due to the force of gravity. Assuming that the ice is released from rest and ignoring friction and air resistance, we can use the conservation of energy to find the expression for the speed of the ice at the bottom of the ramp. The potential energy (PE) at the top will be completely converted into kinetic energy (KE) at the bottom, where PE = mgh and KE = \(\frac{1}{2}mv^2\). Setting these equal gives us the equation mgh = \(\frac{1}{2}mv^2\), which simplifies to v = \(\sqrt{2gh}\).

For the specific example where the ice is released at a height of 0.22 meters (h = 0.22 m) and the ramp is tilted at 40 degrees (which is irrelevant since the angle doesn't affect the final speed in the absence of friction), we can calculate the speed using v = \sqrt{2 * 9.81 * 0.22} which results in approximately 2.083 m/s.

Answer 2

Part 1

Here as we know by energy conservation

Initial potential energy + initial kinetic energy = final potential energy + final kinetic energy

now we will have

`mgh_i + (1)/(2)mv_i^2 = mgh_f + (1)/(2)mv_f^2`

now we will plug in all data into the equation

so we will have

`h_i = h`

`v_i = 0`

`h_f = 0`

from above equation now

`mgh = (1)/(2)mv_f^2`

`v_f = √(2gh)`

Part b)

as we know that

`h = 22 cm`

`\theta = 40 degree`

now from the above equation

`v_f = √(2gh)`

`v_f = √(2(9.81)(0.22))`

`v_f = 2.08 m/s`