How Do You Calculate the Final Speed of a Sliding Ice Block on an Incline?

[rodrj183]

Homework Statement

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9° below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

This problem is found in the University Physics With Modern Physics 13th Edition Book. It is problem number 6.28 in Chapter 6.

Homework Equations

Wtot = Ʃ F times d

Wtot = 1/2mv^2 (final K2) - 1/2 mv^2 (initial K1)

The Attempt at a Solution

Hello guys! I hope everyone is well! There is something I'm not really quite understanding about this

particular problem. Heres my attempt at the solution:

First I found the Work Total done.

I went ahead and found the parallel component of the ice blocks weight to the slope and multiplied it

by the distance of 0.750m down the slope. This gave me a Work Total of 11.75J

I then used the formula Wtot = 1/2mv^2 (final) - 1/2mv^2 (initial) with K1 being 0 since the ice block starts from rest.

11.75J = 1/2(2kg)v^2 - 0

divided 11.75J by 1kg to give me 11.75 m^2/s^2

took sqrt of 11.75 m^2/s^2

V=√11.75 m^2/s^2 = 3.42 m/s

The answer in the book is 2.97m/s.

Im not sure where I went wrong exactly, and I am having a hard time trying to figure out the correct solution. I would love some help! Thanks!

 

[Kishlay]

work done by gravity = change in kinetic energy...!
mgh=.5mv²
h is height of the wedge...!

 

[iRaid]

Just use the fact that:
KE_{i}+PE_{i}=KE_{f}+PE_{f} \implies \frac{1}{2}mv^{2}_{i}+mgh_{i}=\frac{1}{2}mv^{2}_{f}+mgh_{f}
The initial KE and final PE are going to be what?? Then solve from there.

Edit: The trick here is the h isn't the length it slides. Draw a triangle with the hypotenuse of the length it slides and theta being the angle given, then find the height in the y direction.

 

[rodrj183]

I got the answer now, but how come the component of weight parallel to the slope wasn't a valid way of finding the solution? Is it because gravity was the only force doing the work, and gravity can only displace vertically up and down?

 

[rodrj183]

Just use the fact that:
KE_{i}+PE_{i}=KE_{f}+PE_{f} \implies \frac{1}{2}mv^{2}_{i}+mgh_{i}=\frac{1}{2}mv^{2}_{f}+mgh_{f}
The initial KE and final PE are going to be what?? Then solve from there.

Edit: The trick here is the h isn't the length it slides. Draw a triangle with the hypotenuse of the length it slides and theta being the angle given, then find the height in the y direction.

I just actually realized something. The chapter that's in my book doesn't cover GPE yet so the only formulas I was introduced to was the following. The book never mentioned anything about MGH in this chapter. The next chapter mentions it.

Work = Fs cosθ

Wtot = Ʃ F s

and

Wtot = 1/2mv^2 (final) - 1/2mv^2 (initial)

 

[haruspex]

I got the answer now, but how come the component of weight parallel to the slope wasn't a valid way of finding the solution?

Your method was valid, but your numbers look wrong. Did you take cos instead of sine?

 

[rodrj183]

Your method was valid, but your numbers look wrong. Did you take cos instead of sine?

Yep that's exactly what I did.