A 2.20 kg pendulum starts from a height of 5.00m. It swings back and forth through one whole oscillation but only returns to a maximum height of 4.75m. How much negative work was done on the pendulum during the first entire oscillation?
the "negative work" is equal to the lost potential energy due to the height difference
5.00 m g - 4.75 m g
To calculate the negative work done on the pendulum during the first entire oscillation, we need to use the concept of mechanical energy and the work-energy principle.
The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE). In this case, the potential energy of the pendulum is given by its gravitational potential energy, which is equal to mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
At the beginning of the oscillation, the pendulum is at a height of 5.00m, so its initial potential energy is mgh. At the maximum height it reaches during the oscillation, which is 4.75m, its potential energy is mgh'.
Since there is no significant loss of energy due to factors like friction or air resistance, the total mechanical energy remains constant throughout the oscillation. Therefore, the negative work done on the pendulum is equal to the change in potential energy from the initial to the final state.
Negative work can be represented as -W and is given by the equation:
-W = ΔPE = mgh - mgh'
Now, let's plug in the given values:
mass of the pendulum, m: 2.20 kg
acceleration due to gravity, g: 9.8 m/s^2
initial height, h: 5.00 m
final height, h': 4.75 m
-W = (2.20 kg)(9.8 m/s^2)(5.00 m) - (2.20 kg)(9.8 m/s^2)(4.75 m)
Calculating this expression will give you the magnitude of the negative work done on the pendulum during the first entire oscillation.