A 800kg , light-weight helicopter ascends from the ground with an acceleration of 3.00m/s2 .
Part A
Over a 6.50s interval, what is the work done by the lifting force?
Part B
What is the work done by the gravitational force?
Part C
What is the work done by the net force on the helicopter?
work=mgh/time=mg/time * 1/2 a time^2
work=1/2 mga*time
To find the work done by a force, we use the formula:
Work = Force x Distance x cos(θ)
Where:
- Work is the amount of energy transferred by the force,
- Force is the magnitude of the force applied,
- Distance is the displacement of the object in the direction of the force,
- θ is the angle between the force and the direction of displacement.
Part A: Work done by the lifting force.
The lifting force is responsible for the helicopter's ascent. Since it is moving vertically upward, the force and displacement are in the same direction. Therefore, the angle θ between them is 0 degrees, and cos(0) = 1.
We need to find the magnitude of the lifting force first. To do this, we can use Newton's second law:
Force = mass x acceleration
Given:
- Mass of the helicopter (m) = 800 kg
- Acceleration (a) = 3.00 m/s^2
Plug in the values:
Force = 800 kg x 3.00 m/s^2 = 2400 N
The distance traveled by the helicopter is not given directly, but we can use the kinematic equation:
Distance = (1/2) x acceleration x time^2
Given:
- Time (t) = 6.50 s
Plug in the values:
Distance = (1/2) x 3.00 m/s^2 x (6.50 s)^2 = 67.125 m
Now we can calculate the work done by the lifting force:
Work = 2400 N x 67.125 m x cos(0) = 160,800 J
Therefore, the work done by the lifting force over the 6.50s interval is 160,800 J.
Part B: Work done by the gravitational force.
Since the helicopter is ascending vertically, the gravitational force is acting in the opposite direction of the displacement. Therefore, the angle θ between them is 180 degrees, and cos(180) = -1.
The magnitude of the gravitational force can be found using the formula:
Force = mass x acceleration due to gravity
Given:
- Mass of the helicopter (m) = 800 kg
- Acceleration due to gravity (g) = 9.81 m/s^2
Plug in the values:
Force = 800 kg x 9.81 m/s^2 = 7,848 N
The distance traveled is the same as in Part A:
Distance = 67.125 m
Now we can calculate the work done by the gravitational force:
Work = 7,848 N x 67.125 m x cos(180) = -526,995 J
Therefore, the work done by the gravitational force over the 6.50s interval is -526,995 J.
Part C: Work done by the net force on the helicopter.
The net force is the vector sum of the lifting force and the gravitational force. Since these forces are in opposite directions, the net force is:
Net Force = Lifting force - Gravitational force
Net Force = 2400 N - 7,848 N = -5,448 N
Using the same distance as before (67.125 m), we can calculate the work done by the net force:
Work = -5,448 N x 67.125 m x cos(θ)
The angle θ in this case can vary depending on the direction of the net force. Since it is not specified, we will assume it to be 0 degrees (cos(0) = 1).
Work = -5,448 N x 67.125 m x cos(0) = -364,944 J
Therefore, the work done by the net force on the helicopter over the 6.50s interval is -364,944 J. Note that the negative sign indicates the direction of the work done (opposite to the displacement).