Calculating Kinetic and Potential Energy of a Moving Object

[janed123]

Homework Statement

Drew kicks a ball (mass=0.250 kg) off his back porch at h= 1.50 m above the ground with an initial speed v= 17.0 m/s at some known angle. Find the initial kinetic energy of the ball. Find the initial potential energy of the ball. Find the kinetic energy and the speed v of the ball when it reaches the top of the ballistic arc, at h = 9.50 m. Find the speed of the ball v when it reaches the ground h=0.

Homework Equations

The Attempt at a Solution

Initial kinetic energy:
KE = 1/2 mv ^2
KE = 1/2 (0.250) (17.0) = 36. 125 J

Initial potential energy:
PE = mgh
PE= (0.250) (9.81) (1.50) = 3.679 J

This is where I got stuck. If I do not know the speed of the ball when it reaches the top of the arc, how am I supposed to find the KE?
I tried to solve for v by doing:
1/2mv^2 + mgh = 1/2mv^2 + mgh

v^2 = 2(9.81) (9.50)
took the square root..and got v = 13.65 m/s
Then i plugged this speed into the KE equation to solve for KE.
Is this correct? It doesn't seem right to me. I remember hearing that at the top of an arc v would equal zero, but I don't know if that only applies with projectile motion.

Help would be appreciated! Thanks

 

[kushan]

At the top of the arc Vertical component of velocity is zero , not the horizontal :D

 

[janed123]

Thanks! And for the last part of the problem at h=0 would the final speed be the same as the initial speed?

 

[Chemicist]

For the initial kinetic equation, do not forget to square your velocity.

 

[Chemicist]

Oh and if I remember correctly, when the ball is at the top of the arc, its velocity is zero. The kinetic energy is transferred all into gravitational energy. Gravitational energy is a type of potential energy.

But please don't take my word for it until someone confirms my statement.

 

[janed123]

Chemicist: I thought it would equal zero too that is where I am still confused...

 

[Chemicist]

Yes it would equal zero. Because in a system, no energy is created or destroyed. So the kinetic energy, or the energy of motion, is converted into pure gravitational energy (potential) at its peak height, which is the top of the arc.

 

[kushan]

if the velocity is zero at top what makes the ball go forward in horizontal direction?

 

[janed123]

So, if v=0 then ke would = 0 as well because the energy is converted into gravitational energy?

 

[Chemicist]

So, if v=0 then ke would = 0 as well because the energy is converted into gravitational energy?

I believe so, but again, don't turn this in until someone can confirm my statement. I don't want to be giving you wrong answers and if I am, I want someone to correct them.

 

[Chemicist]

if the velocity is zero at top what makes the ball go forward in horizontal direction?

I don't believe it moves in a horizontal direction because the energy is converted into gravitational, which is the energy amassed by a height. So once it's at its peak, it has 100% gravitational energy and the force of gravity will push the ball downward.

 

[kushan]

According to you it will push downward but ball moves forward as well as downward it follows a trajectory , the thing which you are telling happens when you throw a ball straight in air Normal to the plane

 

[Chemicist]

According to you it will push downward but ball moves forward as well as downward it follows a trajectory , the thing which you are telling happens when you throw a ball straight in air Normal to the plane

I am just trying to answer the original question. The question you're asking is separate. Regarding projectile motion, which you can read about here:

http://www.physicsclassroom.com/class/vectors/u3l2e.cfm

 

[kushan]

Is the angle given ?

 

[kushan]

http://ncertbooks.prashanthellina.com/class_11.Physics.PhysicsPartI/ch-4.pdf




14 page in projectile motion

 

[kushan]

http://ncertbooks.prashanthellina.com/class_11.Physics.PhysicsPartI/ch-4.pdf




14 page in projectile motion

 

[kushan]

http://i.imgur.com/HBlwr.jpg

At the top vy is zero not vx

 

[ehild]

Oh and if I remember correctly, when the ball is at the top of the arc, its velocity is zero. The kinetic energy is transferred all into gravitational energy. Gravitational energy is a type of potential energy.

But please don't take my word for it until someone confirms my statement.

Chemicist, your statement is not right.

The ball is kicked at an angle with the horizontal:

Homework Statement

Drew kicks a ball (mass=0.250 kg) off his back porch at h= 1.50 m above the ground with an initial speed v= 17.0 m/s at some known angle.

Initially the ball has both horizontal and vertical velocity components. When in air, the only force acting on the ball is vertical. So the ball keeps its initial horizontal velocity component and the vertical component will change according to v(vertical) =V0(vertical)-gt. At the top of the arc, the vertical velocity component is zero, not the KE.

ehild

 

[kushan]

Apply conservation of energy at top taking initial speed v (which is given ) and at the maxima , velocity will be vcos¢ , so you have equation in one variable from where you can find the angle

 

[ehild]

Kushan is right, both the initial speed and height of the ball are given, so you can get the total energy, E, which is conserved. At the top of the arc, the potential energy corresponds to the height of 9.5 m. Subtract it from the total energy, you get the KE there.
Reaching the ground, all energy is kinetic.

ehild

 

[janed123]

Ok I tried redoing the whole problem...

KE= 36.125 J
PE = 0 J

I found the total E to be 36.125 J

Then i found the PE2 to be 23.31 J (0.250) (9.81) (9.50) = 23.31
Then subtracted 23.31 from the total E 36.125 - 23.31 = 12.815
So the KE2 = 12.815 J

Then I found the speed by calculating (17.0)^2-2(9.81)(9.5)
V= 10.13 m/s

KE1 + PE1 = KE2 + PE2
36.125 = 23.31 + 12.815

Does this seem correct?

I also found the speed of the ball v when it reaches the ground to be:
v^2= 2gh
v^2= 2 (9.81) (9.50)
v = 13.65 m/s

 

[ehild]

The ball is kicked from 1.5 m height and reaches 9.5 m above the ground. The initial potential energy is not zero. Redo the calculations. ehild

 

[kushan]

chemicist is even wrong that at arc velocity will be zero
its Horizontal component of Velocity that is NOT ZERO
VERTICAL COMPONENT IS ZERO

 

[kushan]

potential energy is due to the the stand still position of the objects or things.

really? if in a space no force is acting on a obbject and it standing still does it have a potential to do work?

 

[kushan]

yea zero potential

 

[kushan]

so if an object is moving it cannot have potential energy .
potential enrgy is energy due to configuration of an object in an external field

 

[kushan]

and it is a relative concept , that is why you take potential at infinity to be zero , it is a convention :)

 

[kushan]

you can take potential zero anywhere you wish , but for ease of calculation infinity works for most of the cases .

 

[kushan]

lol , if suppose (it is impossible though but still let me contradict you )
there are three altitudes H1 , H2 , H3 and a charged particle is moving upwards , (remeber moving , which means zero potential according to you )

then all three altitudes are equi-potential surfaces.
And we all know no work is done when object ravels through one equipotential surface to other . That means no work done , but that is wrong , now you guess why ?

 

[kushan]

in the above figure gravity is acting downwards

 

[Chemicist]

Thanks for correcting me ehild!

 

[ehild]

The solution is almost correct, Janed. But it is convenient to take the potential energy zero at the ground.

Ok I tried redoing the whole problem...

KE= 36.125 J
PE = 0 J

The ball was kicked from 1.5 m, so the initial potential energy is (0.250) (9.81) (1.50) instead of zero.
Your way of solution is correct otherwise, just correct the total energy according to the initial potential energy.

ehild

 

[ehild]

ya if an object is moving means it does not have any P.E energy

It is wrong, Thinkhigh. When you drop a stone, for example, the velocity increases with time t as v=gt, and its height decreases as h=h₀-0.5gt². The potential energy with respect to the ground is PE=mgh, The kinetic energy is KE=0.5mv². The PE continuously transforms into KE. Substituting v=gt and h=h₀-0.5gt² into the equations for PE and KE, the total energy is

E=0.5mg²t²+mg(h₀-0.5g²t²)=mgh₀, constant, equal to the initial PE.

ehild