Question

As a box slides down a ramp, friction does 23.0J of work. At the bottom of the ramp, the box has 3.8J of kinetic energy and is traveling at 2.8 m/sec. How high is the ramp?

Answer 1

@SolomonZelman

Answer 2

Weird problem...

Answer 3

I just do not have any formulas for this, and need some help setting it up

Answer 4

Yeah working through it now, just dont want to give it away lol

Answer 5

O, thanks!

Answer 6

Hmm it's weird, I'm missing 1 variable...gimme a sec :)

Maybe typing it out here will help

Answer 7

So first, lets look at the fact they gave us the kinetic energy at the bottom of the ramp

3.8J when traveling at 2.8m/s

Remember our equation for kinetic energy?

\[\large K.E = \frac{1}{2}mv^2\]

Since we are given K.E and 'v' we can solve here for the mass of the box...what would that be?

Answer 8

You there?

Answer 9

Regardless, this is just a conservation of energy problem, the potential energy at the top is equal to the kinetic energy at the bottom plus any work done by friction

\[\large P.E = K.E + W_f\]

After calculating your mass, we write out this equation *with the fact we are given both the kinetic energy and the frictional work

\[\large mgh = 3.8J + 23J\]

Solving for 'h' we have

\[\large h = \frac{26.8J}{mg}\]

Plug in your found mass, multiply it by gravity and carry out this equation to find your height!

Answer 10

Ok, sounds good. I'll show my work in a sec!

Answer 11

Thanks, @johnweldon1993

Answer 12

No problem!

Answer 13

@johnweldon1993 I'm sorry, right as you offered this great explanation, I became occupied with something else, and am just now getting back to this. Solving Kinetic Energy for mass gives me a number in Newton meters. What am I doing wrong?

Answer 14

I understand what is going on, but I am messing up on that specific calculation

Answer 15

Newton*meter is not a unit of mass, but of work done.

Answer 16

Sorry, was away for a bit

So the kinetic energy is the problem here?

\[\large K.E = \frac{1}{2}\text{mass}v^2\]
\[\large 3.8J = \frac{1}{2}\text{mass}(2.8(\frac{m}{s})^2)\]
\[\large 3.8J = \text{mass}(3.92\frac{m^2}{s^2})\]
\[\large \text{mass} = \frac{0.97J}{\frac{m^2}{s^2}}\]

We know \(\large J = kg\frac{m^2}{s^2}\) so we have

\[\large \frac{0.97kg\cancel{\frac{m^2}{s^2}}}{\cancel{\frac{m^2}{s^2}}}\]

Leaving \(\large 0.97kg\)

Answer 17

Ah, I see it now. Thanks so much

Answer 18

And the height of the ramp would be ~2.82 meters high.

Answer 19

Sound about right??

Answer 20

Sorry again >.< yeah that's about what I remember getting!