A crate (weight 381 N) is lowered vertically a distance of 3.0 m by two ropes at a constant speed of 1.2 m/s. The tension in rope A is TA = 248 N; the rope makes an angle of ƒÆA = 63�‹ with the horizontal. Rope B makes an angle of ƒÆB = 55�‹ with the horizontal. What is the work that rope A does on the crate?
I thought no work was done in a vertical motion because cos90=0. I can determine the x and y components of tensionA but how do I find out its work on the crate?
The vertical component of the tension force in rope A, which is 248 sin 63 Newtons, performs equal to
-248 sin 63 N x 3.0 m
The answer is negative because it is being lowered.
You don't need information about rope B to answer the question. You also don't have to use the mass of the crate. They provide more data that you need.
Because the velocity stays the same, you don't have to consider addtional work needed to create kinetic energy.
To find out the work done by rope A on the crate, you need to calculate the dot product of the force applied by rope A and the displacement of the crate.
Let's break down the components of the force and displacement vectors:
Force by rope A (FA) can be split into two components:
- The vertical component is TA sin(θA), where TA is the tension in rope A and θA is the angle made by rope A with the horizontal.
- The horizontal component is TA cos(θA).
The displacement of the crate is a vertical distance of 3.0 m.
Since the crate is being lowered at a constant speed, the net force acting on it is zero. Therefore, the work done by rope A (and also rope B) on the crate is zero.
This can be explained by understanding that work is defined as the dot product of force and displacement. The dot product between two vectors is zero when the vectors are perpendicular. In this case, the vertical component of force by rope A is perpendicular to the displacement of the crate. Therefore, no work is done in the vertical direction.
So, in conclusion, the work done by rope A on the crate is zero.