– [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.

man this could not have been handier! i have a test about this tomorrow!

you're the best!

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

I love these videos! Keep up the great work! đź™‚

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

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

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

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

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))

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

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?

Wow sir, your explanation was so easy to understand

Thank You

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

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

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.

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

Bruh you sound slow

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

That voice bugged me the whole time

wish I could give you 1000 likes! Thanks!

Thanks

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

last few videos for me before the ap test , thanks

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

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

Appreciated for your reply

Thank you.

Abel Urbina

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

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

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