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Physics 11 Online
OpenStudy (yacoub1993):

A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this high? (The acceleration due to gravity is 9.8 m/s2.) Show your work.

OpenStudy (yacoub1993):

@bahrom7893 help me please

OpenStudy (bahrom7893):

What's the formula for KE?

OpenStudy (yacoub1993):

i dont know bro

OpenStudy (bahrom7893):

Look it up on Google and post it here. I'm here to guide you, not to give away the solution.

OpenStudy (yacoub1993):

ok wait for me please

OpenStudy (yacoub1993):

is this the formula KE= 1/2 (M X V^2),

OpenStudy (yacoub1993):

@bahrom7893 wat does the formula stand for.

OpenStudy (bahrom7893):

http://www.physicsclassroom.com/class/energy/u5l1c.cfm That explains what the formula stands for.

OpenStudy (yacoub1993):

1/2 * 7.2 * 9.8^2 345.744 is the correct answer

OpenStudy (bahrom7893):

Why? You were given that acceleration is 9.8 m/s^2. The formula stated that you need to multiply by velocity squared.

OpenStudy (yacoub1993):

how should i multiply

OpenStudy (bahrom7893):

Find out velocity from the information they gave you.

OpenStudy (yacoub1993):

or is it like this (1/2) * (7.2) *(9.8^2)^2 33205.25376

OpenStudy (bahrom7893):

Nooo it is squared only once, but it should not be 9.8^2. V is not 9.8

OpenStudy (yacoub1993):

is the velocity 1.2m

OpenStudy (bahrom7893):

Do you know what velocity means?

OpenStudy (yacoub1993):

no let me check wat it means

OpenStudy (yacoub1993):

the rate at which an object changes its position is velocity. RIGHT

OpenStudy (bahrom7893):

Okay, you found the definition somewhere. What exactly does that mean in your own words?

OpenStudy (yacoub1993):

i dont exactly know

OpenStudy (bahrom7893):

Okay, think about speed. If you cover 100km in 1hour, what's your speed?

OpenStudy (yacoub1993):

it is 100km/h

OpenStudy (bahrom7893):

Okay, so your velocity would also be 100km/h. Now imagine traveling backwards with the same speed. What is your velocity?

OpenStudy (yacoub1993):

i think it is the same or 0km/h

OpenStudy (yacoub1993):

it is the same 100km/h

OpenStudy (bahrom7893):

Uhm close, actually since you're traveling backwards, the sign of velocity would be negative. So anything that moves forward, or to the right, or up would have a positive sign, and anything that moves backwards, left, or down would be negative. So if you're traveling forward with 100km/h speed, your velocity is 100km/h. When you're traveling backwards with the same speed, then your velocity is -100km/h.

OpenStudy (yacoub1993):

owk and for the question wat could be the velocity

OpenStudy (bahrom7893):

How did you figure out the speed when I gave you distance and time?

OpenStudy (yacoub1993):

i think u will divide distance/time

OpenStudy (bahrom7893):

You have the distance, but not the time. But we also have acceleration. Can you figure out time from the acceleration?

OpenStudy (yacoub1993):

no i cannot figure it out give me some ideas

OpenStudy (yacoub1993):

or is it like this time=distance/speed

OpenStudy (yacoub1993):

@dumbcow can u please help me

OpenStudy (dumbcow):

ok sorry was reading everything you are given acceleration, im not sure how you determine inital velocity i think you could also use KE = Force*distance KE = (mass*acc) * distance KE = (7.2*9.8)*1.2

OpenStudy (yacoub1993):

@dumbcow is it acceleration or gravity the 9.8

OpenStudy (ybarrap):

Note that energy is conserved, meaning that total energy = potential energy + kinetic energy. So you know kinetic energy (given above). When the dog starts jumping, his potential energy is zero, but his KE is maximum. That means when the dog reaches the peak of his jump, his KE will be zero. Since energy is conserved, the entire energy must be in the potential energy. Potential energy is mass * gravity * height. So equate KE with PE and that is how much KE is needed.

OpenStudy (dumbcow):

its acceleration of gravity

OpenStudy (ybarrap):

Acceleration is gravity in this problem.

OpenStudy (dumbcow):

like @ybarrap said....the Force needed to get off the ground equals the PE of m*g

OpenStudy (ybarrap):

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