Calculating Mechanical Energy of a Launched Ball

[Pamela Pepper]

Hi Everyone,
I can't seem to figure this problem out. Please help!

1. Homework Statement
A ball was launched upwards and vertically at a speed 3 m/second up to a height 4m. Calculate the mechanical energy of the ball if its weight is 5 Newtons and has a mass 0.5kg .

Homework Equations

potential energy= mass x gravitational acceleration x height
kinetic energy= 1/2 mass x velocity x velocity
mechanical energy= potential energy + kinetic energy

The Attempt at a Solution

mechanical energy = potential energy at maximum height
maximum potential energy =mass x gravitational acceleration x maximum height
= 0.5 x 10 x 4
= 20 joules
So, mechanical energy equals 20 joules?
Does this question seem right to everyone?
Many thanks,
Pamela Pepper

 

[BvU]

Hello PP,

You want to sort out the given information, so that you obtain a complete and solvable problem statement.
In your attempt you mention 'maximum height', but: do you know that that maximum is 4 m ?

To me the scenario is a bit unclear: is the 3 m/s at 0 m, or at 4 m height ? My impression is: the latter.

 

[mfb]

You should use units.

The problem statement is weird. If the ball is launched with 3 m/s at ground level and in free fall afterwards it won't reach a height of 4 m. So where does it have 3 m/s and how are you supposed to take this into account?

 

[CGandC]

I assume they want you to calculate it's mechanical energy when it's exactly reached to a height of h=4m,

You can see that they have given you that the mass is moving with initial velocity $v_0$ of 3m/s upwards .

In addition , the only force acting on the mass is gravity , therefore your mass is constrained to move under constant acceleration , so you can solve the problem by using the equation of motion under constant acceleration $y(t) = y_0 + v_0*t + (a/2)(t)^2$ and the equation of mass's velocity in the y direction ( which is the derivative of the location of the particle in the y-direction w.r.t time )
where $y_0$ is the initial height ( it depends on where your coordinate system is placed ) , $v_0$ is the initial velocity in the y-direction , and $a$ is the constant acceleration.
I won't tell you have to solve the problem, but some hints are here.

 

[Apashanka]

vertically at a speed 3 m/second up to a height 4m. Calculate the mechanical energy of the ball if its weight is 5 Newtons and has a mass 0.5kg .

The mechanical energy is always constt. which is equal to KE +PE ,as the ball moves upward the KE decreases and the PE increases from the the base level.
So at the time of launch e.g at the base level PE is 0 if we take it to be reference level.
So the total mechanical energy is KE at the time of launch which remains constt
throughout it's motion.

 

[mfb]

The mechanical energy is always constt. which is equal to KE +PE ,as the ball moves upward the KE decreases and the PE increases from the the base level.
So at the time of launch e.g at the base level PE is 0 if we take it to be reference level.
So the total mechanical energy is KE at the time of launch which remains constt
throughout it's motion.

See my previous post: It doesn't have enough kinetic energy to reach the height given in the problem statement. Something else has to be there.

OP didn't come back to clarify the problem statement, this might stay a mystery.

 

[Apashanka]

See my previous post: It doesn't have enough kinetic energy to reach the height given in the problem statement. Something else has to be there.

OP didn't come back to clarify the problem statement, this might stay a mystery.

Oh that means the total energy (KE) at the time of launch is less than the total energy at heighest point.

 

[mfb]

Yes, that suggests there is some additional source of energy we don't know about. Does it keep its speed? Does something else happen?
I don't think it helps to speculate, we need a reply from OP, and as long as we don't have that we can just let the thread rest.