Question

What minimum speed does a 100 g puck need to make it to the top of a 4.60 m -long, 24.0 ∘ frictionless ramp?

Answer 1

Answer:

6.05 m/s

Explanation:

In order for the puck to reach the top of the ramp, its initial kinetic energy must be equal to its final potential energy.

So we can write:

$KE=PE\\\frac{1}{2}mv^2 = mgh$ (1)

where

m = 100 g = 0.1 kg is the mass of the puck

v is the initial speed of the puck

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the height of the ramp

Here we know that

d = 4.60 m is the length of the ramp

$\theta=24.0^{\circ}$ is the angle of the ramp

So its height is

$h=d sin \theta = (4.60)(sin 24.0^{\circ})=1.87 m$

So now we can re-arrange eq (1) to find the minimum speed of the puck:

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.87)}=6.05 m/s$