Question

An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a height h(t) = -4.9t 2 + 24t + 60 (in meters). The object has mass m = 2 kilograms. The kinetic energy of the object is given by K = __1 2mv2 , and the potential energy is given by U = 9.8mh. Find an expression for the total kinetic and potential energy K + U as a function of time. What does this expression tell you about the energy of the falling object?

Answer 1

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

$K=\frac{1}{2}mv^{2}$

For the object thrown in the air:

$K=\frac{1}{2}.2.[v(t)]^{2}$

$K=(-9.8t+24)^{2}$

$K=96.04t^{2}-470.4t+576$

Kinetic energy of the object as a function of time: $K=96.04t^{2}-470.4t+576$

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

$U=mgh$

For the object thrown in the air:

$U=9.8.2.h(t)$

$U=9.8.2.(-4.9t^{2}+24t+60)$

$U=-96.04t^{2}+470.4t+1176$

Potential energy as function of time: $U=-96.04t^{2}+470.4t+1176$

Total kinetic and potential energy, also known as mechanical energy is

TME = $96.04t^{2}-470.4t+576$ + ($-96.04t^{2}+470.4t+1176$)

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.