A 230 kg crate hangs from the end of a rope of length L=12m. You push horizontal

Question

A 230 kg crate hangs from the end of a rope of length L=12m. You push horizontally on the crate with a varying force F to move it distance d=4m to the side. a) What is the magnitude of the F when the crate is in this final position? during the crate's displacement, what are b) The total work done on it? c) The work done by the gravitational force on the crate? d) The work done by the pull on the crate from the rope? 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? f) Why is the work of your force not equal to the product of the horizontal displacement and the answer 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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