A rescue helicopter lifts a 70 kg person straight up by means of a cable. The person has an
upward acceleration of 0.70 m/s2
and is lifted from rest through a distance of 9 m.
What is the tension in the cable?
How much work is done by the tension in the cable?
How much work is done by the person's weight?
Use the work-energy theorem and find the final speed of the person.
To find the tension in the cable, we need to consider the forces acting on the person when they are being lifted. In this case, there are two forces at play:
1. The person's weight (mg), acting downward.
2. The tension in the cable, acting upward.
Let's start with the equation of motion for the person:
F_net = ma
Since the person is being lifted vertically, we can write the equation as follows:
T - mg = ma
Now, substitute the given values:
T - (70 kg)(9.8 m/s^2) = (70 kg)(0.70 m/s^2)
Calculate the tension:
T = (70 kg)(0.70 m/s^2) + (70 kg)(9.8 m/s^2) = 686 N
Therefore, the tension in the cable is 686 N.
Next, let's calculate the work done by the tension in the cable. The work done by a force is given by:
Work = Force × Distance × cos(θ)
In this case, the force is the tension in the cable (686 N) and the distance is 9 m. Since the force and the displacement are in the same direction (straight up), the angle (θ) between them is 0 degrees (cos 0 = 1).
So, the work done by the tension in the cable is:
Work = (686 N)(9 m)(cos 0) = 6174 Joules
The work done by the person's weight can be calculated using the same formula, but the angle (θ) between the force and displacement is 180 degrees (cos 180 = -1), as the displacement is in the opposite direction of the force.
Therefore, the work done by the person's weight is:
Work = (-70 kg)(9.8 m/s^2)(9 m)(cos 180) = -6174 Joules (negative sign indicates work done against the force)
Now, let's use the work-energy theorem to find the final speed of the person. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
In this case, the initial kinetic energy is zero (as the person starts from rest). The work done by the tension in the cable and the work done by the person's weight will contribute to the final kinetic energy.
So, the total work done on the person is:
Total Work = Work done by tension + Work done by weight = 6174 J + (-6174 J) = 0 J (net work is zero)
Since the net work is zero, the change in kinetic energy is also zero. Therefore, the final kinetic energy is zero.
Using the equation for kinetic energy:
KE = (1/2)mv^2
Since the final kinetic energy is zero, we can write:
0 = (1/2)(70 kg)(v^2)
Solving for v:
v^2 = 0 m^2/s^2
v = 0 m/s
Thus, the final speed of the person is 0 m/s (stationary).