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.
In its new equilibrium position, the rope is inclined to vertical by an angle
arcsin theta = 4.9/14 = 20.49 degrees
In the process, the crate is raised a distance
delta Y = 14 m (1 - cos theta) = 0.886 m
1. Rope tension T is given by
T cos theta = Mg
Solve for T
The applied horizontal force F needed to hold it in equilibrium is
F = T sin theta
3. The work done on the crate is the potential energy change due to the higher elevation, M g delta Y. (See the calculation of delta Y above)
2. The work done by the crate is the negative of the previous answer.
fsg
fhs
gyhhh
j
To solve this problem, we can break it down into three parts.
1. Magnitude of the applied force F when the crate is at rest in its final position:
- Since the crate is at rest, the net force on it must be zero. Therefore, the applied force F must balance the weight of the crate.
- The weight of the crate can be calculated using the formula: Weight = mass * acceleration due to gravity.
- In this case, weight = (290 kg) * (9.8 m/s^2) = 2842 N.
- Since the applied force F balances the weight, the magnitude of F is also 2842 N.
2. Work done by the weight of the crate:
- The work done by a force is given by the formula: Work = Force * Distance * cos(angle).
- In this case, the force is the weight of the crate, which acts vertically downward.
- The angle between the weight force and the displacement of the crate is 180 degrees (since the force and displacement are in opposite directions).
- Therefore, the work done by the weight of the crate is: Work = (2842 N) * (4.9 m) * cos(180 degrees) = -13931.8 Joules.
- Note that the negative sign indicates that the work done by the weight is in the opposite direction of the displacement.
3. Work done by you on the crate:
- You are pulling horizontally on the crate in the direction of its displacement.
- Since the force and displacement are in the same direction, the angle between them is 0 degrees.
- Therefore, the work done by you on the crate is: Work = (Force applied by you) * (Displacement) * cos(0 degrees).
- To find the force applied by you, we need to use the equation for static friction: Force applied by you = (Coefficient of static friction) * (Normal force).
- The normal force is equal to the weight of the crate, which is 2842 N.
- The coefficient of static friction can vary depending on the surface, but let's assume it is 0.2 for this example.
- Therefore, the force applied by you is: (0.2) * (2842 N) = 568.4 N.
- Finally, the work done by you on the crate is: Work = (568.4 N) * (4.9 m) * cos(0 degrees) = 2785.16 Joules.
So, the answers to the three parts are:
1. The magnitude of the applied force F when the crate is at rest in its final position is 2842 N.
2. The work done by the weight of the crate is -13931.8 Joules.
3. The work you do on the crate is 2785.16 Joules.