A 810.0 kg pink grapefruit is accelerated horizontally from rest to velocity of 11.0 m/s in 5.00 seconds by horizontal force. How much work is done on the pink grapefruit assume no function?
The final KE must equal the work done to accelerate it.
finalKE=1/2 m vf^2
That is the smart way. The harder way is this:
vf^2=vi^2+2ad
but a=force/mass
vf^2=vi^2+2Force*distance/mass
and of course, force times distance is work done.
Yea, they are essentially the same method.
To find the amount of work done on the pink grapefruit, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
The kinetic energy of an object is given by the equation:
KE = (1/2) mv^2
Where:
KE = kinetic energy
m = mass
v = velocity
First, let's calculate the initial kinetic energy (KE_initial) of the grapefruit when it is at rest:
KE_initial = (1/2) m (0)^2 = 0
Next, let's calculate the final kinetic energy (KE_final) of the grapefruit when it reaches a velocity of 11.0 m/s:
KE_final = (1/2) m (11.0)^2 = (1/2) (810.0) (11.0)^2 = 49,005 J (joules)
Now, we can find the work done on the grapefruit by taking the difference between the final and initial kinetic energies:
Work done = KE_final - KE_initial
= 49,005 J - 0 J
= 49,005 J (joules)
Therefore, the work done on the pink grapefruit is 49,005 Joules.