A dild of mass 23 ka s ings at the end of an olastic cord. At th·bottom of the s

Question

A dild of mass 23 ka s ings at the end of an olastic cord. At th·bottom of the s ing, the child's vai oty is ho izontal, a d the spee is 5 m s. At this instant t e co is 4.5 m la o aks th. +x di ection to beha 20 ta a xl to the ight the y direct tù bé up a and the .Idrectio tobe out of tht page. (A] At th intant. "Nhat is the parallel com penent efthe rate of thang" af the eld's mnmentum? . k m/ss [D} At this instant, what the perpendicular component of the rate of dange of the Chld's momentum? 14. c) At this instant, what is the net ferce acting on the child? (d) what i the magitude cf the force that the elastic cord exerts on the child at helps to draw , ciagam of the forces.) e) The relaxed length of the elastic card is 4.42 ·Wi at is the stirness oftite cord? (Use the exact value you entered in aart(d) to made this calculation.)

Explanation / Answer

a] Rate of change in momentum in x = Fx = 0

because there is no force in x

b] Rate of change in momentum in vertical direction = mv^2/r

= 23*5^2/4.5 = 128 kg m/s^2

c] Net force acting on child = mv^2/r

= 23*5^2/4.5 = 128 N

d] Force due to cord, T

T - mg = mv^2/r

T = mg + mv^2/r

= 23*9.8 + 23*5^2/4.5

= 353 N

e] Comparing with spring, T = kx

k = T/x = 353/[4.50 - 4.42]

= 4412.5 N/m

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