1.) A 4.00kg object is mounted on frictionless, straight, horizontal rails. It i

Question

1.) A 4.00kg object is mounted on frictionless, straight, horizontal rails. It is attached to a cable that exerts a horizontal force in the +x direction. The force in the object's position being a function of time: x(t) = (9.50m/s^2) t^2 + (73.0m/s^2) T^3. The cable is designed to snap at exactly 0.250 seconds. The object starts at a position x=0 and at a time t=0.

a.) What is the object's position when the cable snaps?

b.) What is the speed of the object at the moment the cable snaps?

c.) Write an expression for the cable's force as a function of time, F(t). What is its value at t = .220 seconds?

Please provide a clear step by step solution. Thank you.

Explanation / Answer

a)T=mg = 4*10 =40N

Now x(0.25)=9.5(0.25)^2+73(40)=4.672*10^6m

b)speed = dx/dt= 9.5(2t) =9.5(2*0.25) = 4.75m/s

c)F=mdv/dt = 4(9.5*2) =76 implying F(t) = 76N, constant

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