A helicopter's speed increases
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Newton's second law and kinematics allow finding the answers for the changes in weight when the body stops are:
A) The apparent weight is 3548 N directed upwards
B) The factor of change of weights is 5
Given parameters
- Free fall time t1 = 2.0 s
- The time tot stops at t2 = 0.5 s
- Mass m = 65 kg
To find
A) Apparent weight
B) Factor that exceeds the actual weight
Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.
This exercise we will do them in two parts
1st part. Freefall
We look for the velocity at the end of the trajectory, as part of rest the initial velocity is zero
v = v₀ - g t
Where v is the velocity, vo the initial velocity, g the acceleration due to gravity and y time
v = -gt
v = -9.8 2.0
v = -19.6 m / s
The negative sign indicates that the velocity is directed downwards
2nd part. Braking movement
We look for the acceleration to stop the body, in this case the final velocity is zero
= v + a t
0 = v + at
a = - v / t
a =
a = + 39.2 m / s²
The positive sign indicates that the acceleration is directed upward.
A) Newton's second law states that the force is proportional to the mass and the acceleration of the body
∑ F = m a
F -W = m a
F = m (a + g)
F = 65 (39.2 + 9.8)
F = 3185 N
Body weight is
W = mg
W = 65 9.8
W = 637 N
therefore the apparent weight is
ΔW = F -W
ΔW = 3185 - 637
Δw = 2548 N
B) the weight change factor
Factor = F / W
Factor =
factor = 5
In conclusion, using Newton's second law and kinematics, we can find the answers for the changes in weight when the body stops are:
A) The apparent weight is 3548 N directed upwards
B) The factor of change of weights is 5
Learn more about Newton's second law and kinematics here: