A 2 kg rock is released from rest at a height of 29 m. Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights;20m 15m 9m

* physics - bobpursley, Wednesday, November 17, 2010 at 9:32am

PE= mgh at any height.

KE at any height= InitialPE-finalPE
Total mech energy = KE+ PE

Please Can you explain how I can calculate kinetic energy if I don't know velocity I know formula mv^2/2 in this case I have to know the velocity.

if KE = energy lost in PE

and you fell x distance, then

1/2 mv^2=mg*x and you calculate v from that.

To calculate the kinetic energy (KE) at a specific height without knowing the velocity, you can use the conservation of energy principle. The principle states that the total mechanical energy (TME) of a system remains constant as long as no external forces are acting on it. In this case, the only force acting on the rock is gravity.

The formula to calculate TME is:

TME = KE + PE

Where:
TME is the total mechanical energy
KE is the kinetic energy
PE is the gravitational potential energy

To find the kinetic energy (KE) at a specific height, you need to know the initial potential energy (PE) at the release height. You can calculate the initial PE using the formula:

PE = mgh

Where:
m is the mass of the rock (2 kg)
g is the acceleration due to gravity (9.8 m/s^2)
h is the height from which the rock is released

Now, to find the KE at a specific height, use the formula:

KE = Initial PE - Final PE

Where:
Initial PE is the potential energy at the release height (29 m)
Final PE is the potential energy at a specific height

Substitute the values from the problem and solve the equations for each given height (20 m, 15 m, and 9 m) to find the KE at those heights. Remember, potential energy decreases as height decreases.

Finally, calculate the total mechanical energy (TME) at each height by adding the calculated KE and PE values together. This will give you the total energy of the rock at each respective height.