What minimum speed does a 100 g puck need to make it to the top of a 2.9-m-long, 34° frictionless ramp?

Step No: 1

The diagram to calculate height of the ramp using the length of the ramp and inclination of the ramp is as follows:

A use full trigonometry ratio in a right angle triangle is,

sinθ=hypotonuse/oppositeside

sin34∘=2.9m/h

Solve for h.

h=(2.9m)sin34∘=1.62m

Step No: 2

The expression for kinetic energy K of a moving object of mass m is,

K=1/2mv2

The potential energy of an object which is at a height h above the reference level is,

U=mgh

The total energy of the puck at the bottom of the ramp is its kinetic energy. If this total kinetic energy is converted into total potential energy of the puck when it reaches the top of the ramp.

K=U

Substitute 1/2mv2 for K and mgh for U.

1/2mv2=mgh

1/2v2=gh

Rearrange the equation for v.

v=√2gh

Substitute 9.8m/s2 for g and 1.62 m for h.

v=√2(9.8m/s2)(1.62m)=5.64m/s