Does using (Fcos(theta))d work for this problem or should cos be replaced with sin?
A cheerleader lifts his 66.4 kg partner straight up off the ground a distance of 0.718 m before releasing her.If he does this 27 times, how much work has he done?
1. The weight of the partner is:
F = mg = (66.4 kg)(9.8 N/kg)
2. The work done in a single lift is:
work = F x distance lifted.
3. Total work = (work /lift)(27 lifts)
How do you find the force to plug in
To determine the work done by the cheerleader, we can use the equation:
Work = Force x Distance x cos(theta)
In this case, the force applied is the weight of the partner, which can be calculated using the equation:
Force = mass x gravity
To calculate the force, we need to know the value of gravity, which is approximately 9.8 m/s².
Using the given mass of 66.4 kg and the value of gravity, we can find the force:
Force = 66.4 kg x 9.8 m/s² = 650.72 N
Now that we have the force, we can calculate the work done by multiplying it with the distance and the cosine of the angle theta. In this problem, the distance is given as 0.718 m. However, the angle theta is not provided.
The angle theta represents the angle between the force applied and the direction of the movement. Since the problem does not provide any information regarding the angle, we cannot accurately determine the value of cos(theta) or sin(theta).
Therefore, without knowing the angle theta, we cannot use the equation (Fcos(theta))d or replace cos with sin. The angle must be provided to determine the exact work done by the cheerleader.