A 290- kg crate hangs from the end of a 14.0 m long rope. You pull horizontally with a varying force to move it a distance d = 4.9 m to the right.
1.What is the magnitude of the applied force F when the crate is at rest in its final position?
2.What is the work done by the weight of the crate?
3.What is the work you do on the crate?
I am very much stuck, and I would very much appreciate help on all three parts. Thanks.
This was posted and answered elesewhere.
To solve these questions, we need to consider the forces acting on the crate and apply the laws of physics related to work and equilibrium. Here's how we can find the answers to each question:
1. To determine the magnitude of the applied force F when the crate is at rest in its final position, we need to consider the forces in equilibrium. When the crate is at rest, the forces acting on it are the applied force F, the weight of the crate (mg), and the tension in the rope (T). The tension is equal to the weight of the crate, which can be calculated using the formula T = mg, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, T = (290 kg)(9.8 m/s^2) = 2842 N. Since the crate is at rest, the horizontal component of the applied force F must be equal to the tension in the rope. Thus, the magnitude of the applied force F is 2842 N.
2. To calculate the work done by the weight of the crate, we use the formula W = Fd cos(theta), where W is the work done, F is the force applied, d is the distance moved, and theta is the angle between the force and the displacement. In this case, the weight of the crate acts vertically downwards, and the displacement is horizontal, so the angle theta is 90 degrees and cos(theta) = 0. Therefore, the work done by the weight of the crate is zero.
3. The work done by you on the crate can be calculated using the same formula: W = Fd cos(theta). In this case, the applied force is horizontal, and the displacement is also horizontal, so the angle theta is 0 degrees and cos(theta) = 1. Therefore, the work done by you on the crate is W = Fd. Using the values given in the problem, this is equal to (2842 N)(4.9 m) = 13921.8 Nm or joules.
I hope that helps! Let me know if you have any further questions.