<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
=>
=> =
Now substituting thje values
=> =
=>
=>
=>
Now the tension in the string resolve into components
The vertical component supports the weight
=>
=>
=>
=>
=>
=>T =2391N
Therefore the horizontal force
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
=
=
=
=
=
= -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)