A plane designed for vertical takeoff has a mass of 51 kg. Find the net work Wnet as it accelerates upward at 1 m/s2 through a distance of 34 m after starting from rest.
vf^2=2ad
1/2 m vf^2=mad
so total work = mgh+mah
To find the net work, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy.
The formula for net work is given by:
Wnet = ΔKE
Where:
Wnet is the net work done on the object
ΔKE is the change in kinetic energy
The change in kinetic energy can be calculated using the formula:
ΔKE = (1/2) * m * (vf^2 - vi^2)
Where:
m is the mass of the object
vf is the final velocity
vi is the initial velocity
Given:
mass, m = 51 kg
acceleration, a = 1 m/s^2
distance, d = 34 m
initial velocity, vi = 0 m/s (as it starts from rest)
To find the final velocity, we can use the kinematic equation:
vf^2 = vi^2 + 2 * a * d
Substituting the given values into the equation:
vf^2 = 0^2 + 2 * 1 * 34
vf^2 = 68
vf = √68
vf ≈ 8.246 m/s (rounded to three decimal places)
Now we can calculate the change in kinetic energy:
ΔKE = (1/2) * m * (vf^2 - vi^2)
ΔKE = (1/2) * 51 * (8.246^2 - 0^2)
ΔKE ≈ 1063.289 J (rounded to three decimal places)
Finally, we can find the net work:
Wnet = ΔKE
Wnet ≈ 1063.289 J
Therefore, the net work done on the plane is approximately 1063.289 J.
To find the net work done by the plane as it accelerates upward, we need to use the work-energy principle. The work done, W, is equal to the change in kinetic energy, ΔKE.
The formula for work is given by:
W = ΔKE
The change in kinetic energy is calculated using the formula:
ΔKE = (1/2) * mass * (final velocity^2 - initial velocity^2)
Here, the plane starts from rest, so the initial velocity (vi) is 0.
ΔKE = (1/2) * mass * (vf^2 - 0^2)
Given:
mass (m) = 51 kg
acceleration (a) = 1 m/s^2
distance (d) = 34 m
To find the final velocity (vf), we can use the following kinematic equation:
vf^2 = vi^2 + 2ad
Since the plane starts from rest, vi = 0.
vf^2 = 0^2 + 2 * a * d
vf^2 = 2 * 1 * 34
vf^2 = 68
vf = √68 (taking square root on both sides)
vf ≈ 8.2462 m/s
Now, substitute the values of mass and vf in the equation for ΔKE:
ΔKE = (1/2) * 51 kg * (8.2462 m/s)^2
ΔKE ≈ (1/2) * 51 * 67.965 m^2/s^2
ΔKE ≈ 1726.8925 J
Therefore, the net work done by the plane as it accelerates upwards is approximately 1726.8925 Joules (J).