A cup of mass m, placed on the edge of a table of height h, has potential energy mgh, where g is the acceleration due to gravity. Someone slams a door, causing the cup to fall to the floor. If h = 100 cm, what is the approximate speed of the cup when it hits the floor?
(Note: Answer: 4.2 m/s. Show the steps in your solution. For such a short fall, you may neglect air resistance. To simplify the numerical part of the calculation, you may use the following approximations: g ! 9.0 m/s/s and "2 ! 1.4)
The kinetic energy will equal mgh
1/2 mv^2=mgh
v= sqrt(2gh) where h is in meters.
I am somewhat surprised that this question was given in a physics class. Who in the world would approximate g as 9m/s^2? Silly, silly, silly. Why not approximate 1meter as 94cm? or a kilogram as 920grams? or a mile as 5000 feet?, or as done in one state, legally defince PI as 3, instead of 3.14... , after all, it does simplify calculations.
CHEM7300 Assignment?
To find the approximate speed of the cup when it hits the floor, we can use the principle of conservation of energy. We know that the potential energy of the cup at the edge of the table is equal to the kinetic energy of the cup when it hits the floor.
The formula for potential energy is given by:
Potential Energy = mass (m) * acceleration due to gravity (g) * height (h)
So, the potential energy of the cup at the edge of the table is mgh.
Next, we equate the potential energy to the kinetic energy:
Potential Energy = Kinetic Energy
mgh = (1/2) * mass (m) * velocity^2
Since we are interested in finding the velocity when it hits the floor, let's solve for velocity:
mgh = (1/2) * mv^2
Canceling out the mass (m) on both sides:
gh = (1/2) * v^2
v^2 = 2gh
Taking the square root on both sides:
v = sqrt(2gh)
Now, plug in the given values:
g = 9.0 m/s^2 (approximated as 9.0)
h = 100 cm = 1 meter (since 1 cm = 0.01 m)
v = sqrt(2 * 9.0 * 1)
v = sqrt(18)
v ≈ 4.24 m/s (approximated as 4.2 m/s)
Therefore, the approximate speed of the cup when it hits the floor is 4.2 m/s.