A 0.245kg ball is thrown straight up from 2.20m above the ground. Its initial vertical speed is 8.80m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.
Fg = m*g = 0.245kg * 9.8N/kg = 2.401 N.=
Force of gravity.
D = d1 + d2
d1=(V^2-Vo^2)/2g=(0-8.8^2)/-19.6=3.951 m
Up.
d2 = d1 + 2.20 = 3.951 + 2.20 = 6.15 m,
Down.
Wg = Fg*(d1+d2)=2.401*(3.951+6.15)=24.3
Joules.
To calculate the total work done by the force of gravity, we can use the work-energy principle. The work done by a force is equal to the change in kinetic energy of an object.
First, let's calculate the initial kinetic energy of the ball when it is thrown straight up. The formula for kinetic energy (KE) is given by KE = 1/2 * mass * velocity^2.
Plugging in the values, we have:
Initial KE = 1/2 * 0.245kg * (8.80m/s)^2
Next, let's calculate the final kinetic energy of the ball just before it hits the ground. Since the ball is at its highest point, its final velocity will be zero.
Final KE = 1/2 * 0.245kg * (0m/s)^2
The change in kinetic energy is given by the difference between the final and initial kinetic energies, which is equal to the work done by the force of gravity.
So, the total work done by the force of gravity is:
Work = Final KE - Initial KE
Substituting the values, we have:
Work = [1/2 * 0.245kg * (0m/s)^2] - [1/2 * 0.245kg * (8.80m/s)^2]
Simplifying the equation, we get:
Work = 0 - [1/2 * 0.245kg * (8.80m/s)^2]
Now we can calculate the answer by plugging in the values and solving the equation.