Indoor, Outdoor & Kids' Trampolines

# Conservation of Energy Problem with Friction, an Incline and a Spring by Billy

– [Voiceover] Good morning. My name is Billy, and this
is my video project for ♫ Flipping Physics ♫ Let’s do an example problem. A block with a mass of 11
grams is used to compress a spring a distance of 3.2 centimeters. The spring constant of the
spring is 14 Newtons per meter. After the block is released, it slides along a level,
frictionless surface until it comes to the bottom
of a 25 degree incline. If the coefficient of kinetic friction between the block and the incline is 0.30, to what maximum height
does the block slide? Let’s see. Our knowns are the spring
constant of the spring is 14 Newtons per meter. The angle of the incline is 25 degrees. The mass of the block is 11 grams. The initial compression of the spring, x, is 3.2 centimeters, and the coefficient of kinetic friction between the incline and the block is 0.30. And we are solving for maximum height, so height maximum equals question mark. Because the spring constant
is in Newtons per meter and a Newton is a kilogram
meter per second squared, we need all our known
values in kilograms, meters, and seconds, the base SI units. In other words, we need to convert the mass of the block
from grams to kilograms, so we multiply 11 grams by
one kilogram by 1,000 grams, and the mass of the
block is 0.011 kilograms. We also need to convert the initial compression
of the spring to meters, so x initial equals 3.2 centimeters times one meter divided
by 100 centimeters, or 0.032 meters. On the level surface, there
is no work done by friction and no work done by the force applied. Therefore, we can use
conservation of mechanical energy, so we need to set the
initial and final points. Let’s set the initial point
where the block is completely compressing the spring and the final point at the base of the incline. Because there is friction on the incline, mechanical energy is not conserved there, and therefore the final point
cannot be on the incline. Oh, and let’s set the horizontal zero line at the center of mass of the block while it is on the level surface. Now we can use conservation
of mechanical energy on the level surface. Mechanical energy initial
equals mechanical energy final, and all the possible
energies which could be there initially and finally are kinetic energy, gravitational potential energy, and elastic potential energy. Therefore, one-half mass
times velocity initial squared plus mass times the
acceleration due to gravity times the initial height plus one-half times the spring constant
times the initial displacement of the spring squared equals one-half mass times
the velocity final squared plus mass times the
acceleration due to gravity times the final height plus one-half times the spring constant
times the final displacement of the spring squared. Now we need to determine
which energies are zero. The height initial and
final are both zero, so the gravitational potential
energy initial and final are both zero. The initial velocity of the block is zero, so the initial kinetic energy is zero, and the block is no longer on
the spring at the final point, so the final elastic
potential energy is zero. Therefore, we are left with one-half times the spring constant
times the initial displacement of the spring squared equals
one-half times the mass of the block times the final
velocity of the block squared. Sadly, everybody did not
bring mass to the party. I guess we can multiply
the whole equation by two. Then divide both sides
by the mass of the block. Then take the square root
of the whole equation to get the final velocity of the block equals the square root of the quantity, the spring constant times
the initial displacement of the spring squared all
dived by the mass of the block. With numbers, that is
the square root of 14 times 0.032 squared divided by 0.011, which equals 1.14161 meters per
second, the final velocity of the block at the end
of the level surface, which is also the initial
velocity of the block on the incline. Now we need to look at the incline. We get to draw a free
body diagram for the block as it goes up the incline. The force of gravity is straight down. The force normal is perpendicular
to the incline and up, and the force of kinetic friction
is parallel to the incline and opposes the motion of the block, which is sliding up the incline, so the force of kinetic
friction is down the incline. Now we can break the force of gravity up into its components, the force of gravity parallel
to and down the incline and the force of gravity
perpendicular to the incline and down. Now we need to sum the forces in the perpendicular direction. The net force in the
perpendicular direction equals the force normal minus the
force of gravity perpendicular, and it also equals mass times acceleration in the perpendicular direction. The acceleration in the
perpendicular direction equals zero because the block isn’t moving
perpendicular to the incline, so the force normal equals the force of gravity perpendicular, which equals mass times the
acceleration due to gravity times the cosine of the incline angle. We can put that equation
into our equation holster. No reason to substitute in numbers. I don’t have a numbers dependency. Now I can sum the forces
in the parallel direction. The net force in the parallel direction equals the negative of the
force of gravity parallel minus the force of kinetic friction. Both of those forces are negative because they are to the left. The net force in the parallel
direction also equals mass times acceleration
in the parallel direction. We have equations for both forces, so we can substitute for the
force of gravity parallel, mass times the acceleration due to gravity times the sine of the incline angle, and for the force of kinetic friction, we can substitute in the
coefficient of kinetic friction times force normal. Oh, from our equation holster, we can substitute for the force normal, mass times the acceleration due to gravity times the cosine of the incline angle, and everybody brought mass to the party. (electronic dance music) ♫ Everybody brought mass ♫ Therefore, we can be equitable and take mass from everyone. We get the acceleration parallel equals the negative of the
acceleration due to gravity times the sine of the incline angle minus the coefficient of kinetic friction times the acceleration due to gravity times the cosine of the incline angle With numbers, that is the negative of 9.81 times the sine of 25 minus 0.3 times 9.81 times the cosine of 25, which works out to be negative 6.81315 meters
per second squared. Now we can use one of the uniformly accelerated motion equations
to solve for the displacement of the block in the parallel direction. Let’s see. Velocity final parallel squared equals velocity initial parallel squared plus two times acceleration parallel times displacement parallel. Because we are talking
about the maximum distance the block travels up the incline, the final velocity of the block is zero. Subtract velocity initial
parallel squared from both sides, and then divide both sides by two times the acceleration parallel
to get the displacement in the parallel direction
equals the negative of the initial velocity in
the parallel direction squared divided by two times the
acceleration parallel. With numbers, that is the
negative of the quantity 1.14161 squared divided by two times negative 6.81315, which is 0.095644 meters. Now we can find the maximum
height using sine of theta equals opposite over hypotenuse. Opposite theta is the maximum height, and the hypotenuse is the displacement in the parallel direction. Multiplied by the displacement
in the parallel direction to get the height maximum
equals the displacement in the parallel direction times the sine of the
incline angle, or 0.095644 times the sine of 25 degrees, which is 0.040421 meters. However, let’s convert it to centimeters. Multiply by 100 centimeters
divided by one meter, and with two significant digits, the answer is 4.0 centimeters, the maximum height to
which the block slides is 4.0 centimeters. Thank you very much for
learning with me today. I enjoyed learning with you.

1. kotzzz9 says:

you're the best!

2. AyalaMrC says:

I hope you are going to use Billy's video project as an example of the harder way of doing this.

3. Hector Villodas says:

I love these videos! Keep up the great work! 🙂

4. TheHammer says:

I paused the video when the problem was given and solved it in the exact same way!

5. Dan Fullerton says:

Showing this to my students to get ready for their energy exam at the end of the week, thanks!

6. deanrule70yahoorules says:

The audio in this video is interesting, sounds like the voice is coming from inside my head. It's almost discomforting, but interesting nonetheless.

7. Flipping Physics says:

Thank you to whomever just posted the Polish subtitle translation for this video! It would be awesome if you could identify yourself so I can give you credit here and on my webpage. email is good. flippingphysics(at)gmail(dot)com

8. salusiwe hlambelo says:

i tried using the W(nc)=ME (f) – ME (i) (in the x-direction or in the parallel plane) to find the hmax. i substituted everything well but i got 6.89 m. Could you explain why this formula is wrong for getting the hmax (which is h (f))

9. Adam Royaie says:

Hi mr p This is really best energy problem I pause the video for solve it but I couldn't thank you belly

10. Cam De Leon says:

Why is the velocity of the block the same at the point of it leaving the spring, as it is entering the ramp because it is frictionless?

11. Sunny shah says:

Wow sir, your explanation was so easy to understand
Thank You

12. colorfulcrayon57 says:

I'm currently taking AP Physics C and your videos are so helpful!

13. Arsonic says:

How would you do the part up the ramp only using energy equations?

14. John Appleseed says:

Thanks so much for this video! I have a test on this tomorrow and this makes me feel a lot more confident. The same video with the energy equations method also helped. Again, thanks.

15. Ipfreely31415 says:

On your free body diagram, why do you not have an opposing force to Fkf

16. Tre Davidson says:

Bruh you sound slow

17. seni A says:

Why didn't you use potential energy when you found the max height?

18. YS r34 says:

That voice bugged me the whole time

19. L& L says:

wish I could give you 1000 likes! Thanks!

20. Nooordin Nordin says:

Thanks

21. Razia Sultana Shushanta says:

I have a question. How come the level ground demonstrates the conservation of energy? When you compressed the spring with the pencil, you applied external force. Since there's an external force applied out of the system, the mechanical energy is not reserved, right? I am confused

22. ofek shaltiel says:

last few videos for me before the ap test , thanks

23. Kimaya Wanjari says:

Billy is a genius!!!!!!!!

24. Abel Ubina says:

Hello Prof Billy,

The WORK OUTPUT is more than double the WORK INPUT. So the theory of thermodynamics is misconception.
A 15 ball billiard pool to be strike by white mother ball

When I computed the work done in pool (billiard) the white mother ball break the 15ball.

F = 30lbs (White Mother Ball); D = 3ft (Distance from the white mother ball to the first ball to strike)

SOLVE FOR WORK INPUT

W = F X D; W = 30lbs x 3ft = 90Ft.lbs

SOLVE FOR WORK OUTPUT

First two ball extreme corner of the billiard ball 2 out of 15 ball

W = F X D; W = 30lbs x 3ft = 90ft.lbs X 2ball = 180 ft. lbs

The remaining ball (13pcs ball) can produce more than 90 ft.lbs

The computation shows that the output work done is higher than the input work done

Thank you.

Abel Urbina

25. Shrikantiah TDR says:

Sir these videos are really good even as Crash Courses and for quick revision purpose. Love from an Indian Jee Aspirant❤

EDIT: I'M SHARING YOUR ENTIRE JEE PLAYLISTS TO MY CLASS GROUP:)

26. I'm The flash says:

Why is the velocity at the final point is equal to the initial velocity as it is entering the lamp.