Question

A 230 kg crate hangs from the end of a rope of length L = 12.0 m. You push horizontally on the crate with a varying force F to move it distance d = 4.00 m to the side. (a) What is the magnitude of 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, and (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 to (a)?

Answer 1

From the given figure

 The vertical angle is determined as follows
                               Sinθ = d / l
                                or θ =  Sin^-1( 4 / 12 ) =19.5^o

^{Now the tension in the string resolve into components}

The vertical component supports the weight
                            TCosθ = 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 W_g =F_s.d = - mgh
                                                    0r W_g = - mg l ( 1 - Cosθ )
                                                              = -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 W_F = - W_g 

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        

Answer 2

m = 200 kg, L = 13.0 m, d = 4.00 m
From the figure the angle θ = sin^-1(d/L) =17.92^o   
a) The force F = mg*tanθ = 634 N

b) there is no change in kinetic energy, so the total workdone is zero.
c) The work done by the gravitational force is

          W_g = - m g L(1 - cosθ) = - 1236 J

d)here the angle is 90

        W = 0  

e) The work done by the force is

      W = -  W_g = 1236 J

Answer 3

Given m = 240 kg, L = 11.0 m, s = 4.00 m,
angle of the rope with vertical θ = sin^-1(s/L) =21.3^o
a) F = Tsinθ, mg = Tcosθ
so F = mgtanθ = 918.1 N
b) since there is no motion at the final position, the total work = 0
c) W_mg = mghcos(180) = -mgL(1 - cosθ) =-1.767*10³ J=-1.767 kJ
d) W_T = T*d*cos(90) = 0
e) W_F = -W_mg - W_T= 1.767*10³ J =1.767 kJ