<u>Answer</u>:
(a) magnitude of F = 797 N
(b)the total work done W = 0
(c)work done by the gravitational force = -1.55 kJ
(d)the work done by the pull = 0
(e) work your force F does on the crate = 1.55 kJ
<u>Explanation</u>:
<u>Given</u>:
Mass of the crate, m = 220 kg
Length of the rope, L = 14.0m
Distance, d = 4.00m
<u>(a) What is the magnitude of F when the crate is in this final position</u>
Let us first determine vertical angle as follows
=>![Sin \theta = \frac{d }{L}](https://tex.z-dn.net/?f=Sin%20%5Ctheta%20%3D%20%5Cfrac%7Bd%20%7D%7BL%7D)
=>
=
Now substituting thje values
=>
=
=>
=> ![\theta = Sin^{-1}(0.333)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20Sin%5E%7B-1%7D%280.333%29)
=> ![\theta = 19.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2019.5%5E%7B%5Ccirc%7D)
Now the tension in the string resolve into components
The vertical component supports the weight
=>![Tcos\theta = mg](https://tex.z-dn.net/?f=Tcos%5Ctheta%20%3D%20mg)
=>![T = \frac{mg}{cos\theta}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7Bmg%7D%7Bcos%5Ctheta%7D)
=>![T = \frac{230 \times 9.8 }{cos(19.5)}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B230%20%5Ctimes%209.8%20%7D%7Bcos%2819.5%29%7D)
=>![T = \frac{2254 }{cos(19.5)}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2254%20%7D%7Bcos%2819.5%29%7D)
=>![T = \frac{2254 }{0.9426}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2254%20%7D%7B0.9426%7D)
=>T =2391N
Therefore the horizontal force
![F = TSin(19.5)](https://tex.z-dn.net/?f=F%20%3D%20TSin%2819.5%29)
F = 797 N
b) The total work done on it
As there is no change in Kinetic energy
The total work done W = 0
<u>c) The work done by the gravitational force on the crate</u>
The work done by gravity
Wg = Fs.d = - mgh
Wg = - mgL ( 1 - Cosθ )
Substituting the values
= ![-230 \times 9.8\times 12 ( 1 - cos(19.5) )](https://tex.z-dn.net/?f=-230%20%5Ctimes%209.8%5Ctimes%2012%20%28%201%20-%20cos%2819.5%29%20%29)
= ![-230 \times 9.8\times 12 ( 1 - 0.9426) )](https://tex.z-dn.net/?f=-230%20%5Ctimes%209.8%5Ctimes%2012%20%28%201%20-%200.9426%29%20%29)
= ![-230 \times 9.8\times 12 (0.0574)](https://tex.z-dn.net/?f=-230%20%5Ctimes%209.8%5Ctimes%2012%20%280.0574%29)
= ![-230 \times 9.8\times 0.6888](https://tex.z-dn.net/?f=-230%20%5Ctimes%209.8%5Ctimes%200.6888)
= ![-230 \times 6.750](https://tex.z-dn.net/?f=-230%20%5Ctimes%206.750)
= -1552.55 J
The work done by gravity = -1.55 kJ
<u>d) the work done by the pull on the crate from the rope</u>
Since the pull is perpendicular to the direction of motion,
The work done = 0
e)Find the work your force F does on the crate.
Work done by the Force on the crate
WF = - Wg
WF = -(-1.55)
WF = 1.55 kJ
<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>
Here the work done by force is not equal to F*d
and it is equal to product of the cos angle and F*d
So, it is not equal to the product of the horizontal displacement and the answer to (a)