Physics Work and Energy Question. Please check my answer.

Question

A 5g apple is thrown straight upward with an initial velocity of 20 m/s @ a point of 1.5m height from ground. 
Ignoring air friction, find:

 

a. Total mechanical energy of the apple at initial position.
b. Total potential energy of the apple at maximum position. 

 

a) 5g = 0.005kg

 

    Kinetic Energy = (1/2)(m)(v)^2

                          = (1/2)(0.005)(20)^2

                          = 1J

 

b) Finding max position, 

    s = (u^2 - v^2)/(2a)

       = (20^2)/(2(9.81))

       = 20.39m

 

    Potential Energy = mgΔh

                            = (0.005)(9.81)(20.39-1.5)

                            = 0.9J

 

Is this correct? It feels wrong. Thank you in advance,

Answer 1

For part A you have ignored the potential energy at the starting position of 1.5 m above ground.

U_i = mgh = (0.005kg) (9.81 m/s²) ((1.5 m) =

 

For part B finding the max position is needed as you do.

But not that way.  You have U² - v²  How can you subtract velocity from energy?

 

Since at the top v_f = o we can take advantage of that and the fact that the up and down problems at symmetrical/mirror images with the falling velocity = initial velocity at the same height.

 

Δx = v_f² / 2a   So you can see you were doing the correct maths with numbers but but not from the right formula.

 

Now for the second part of B

 

U = mgh  that's right, but h is the 20.39 m + 1.5 meters ot less than less than

 

So for a) K = the 1 J , and U = mgh where h is 1.5 m - then total them.

and for b) K = 0 (zero v at top) and U is mgh with h = (20.39 m + 1.5 m)

Answer 2

Haizal,

 

You almost nailed it!  Congratulations, your calculation for the kinetic energy and the max position are dead-on.  You have one misstep in the second line of your potential energy calculation.

 

Δh should be 20.39 + 1.5, not minus, since you started your throw 1.5 meters above the ground you need to add that to the max height to get the total height above the ground.