A 230 kg crate hangs from the end of a rope of length L = 13.0 m. You push horiz

Question

A 230 kg crate hangs from the end of a rope of length L = 13.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.
(a) What is the magnitude of F when the crate is in this final position?
_____ N

(b) During the crate's displacement, what is the total work done on it?
____ J

(c) During the crate's displacement, what is the work done by the weight of the crate?
_____ kJ

(d) During the crate's displacement, what is the work done by the pull on the crate from the rope?
_____ J

(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.
_____ kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?
_____________________________________________________

Explanation / Answer

From the given figure  The vertical angle is determined as follows
                               Sin = d / l
                                or =  Sin-1( 4 / 12 ) =19.5o Now the tension in the string resolve into components The vertical component supports the weight
                            TCos = mg
                                 0r T = 230*9.8 / Cos19.5 = 2391N
Therefore the horizontal force F = TSin19.5
                                       0r F = 797 N

b)  The total work done W = 0 asthere is no change in KE

c)  The work done by gravity Wg =Fs.d = - mgh
                                                    0r Wg = - mg l ( 1 - Cos )
                                                              = -230*9.8*12 ( 1 - Cos19.5 )                                                               = - 1.55 kJ

d)  asthe pull  is perpendicular to the direction ofmotion, the work done = 0

e)  Workdone by the Force on the crate WF = - Wg  f) 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        

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