A box with a mass of 2.69 kg is attached to a spring with spring constant k = 93.8 N/m. The coefficient of kinetic friction between the block and the table is µk = 0.477. The box is displaced from the equilibrium position to the left by 0.257 m. If the block is released from rest, calculate the box's displacement when it rests again.
I've tried a couple different ways but cant seem to quite get it. I keep coming out with a negative answer for some reason.
I've tried Fnet=ma
Fnet=0
Fs1-Ff-Fs2-Ff=0
Fs1-2Ff=Fs2
(kL1-2uK*mg)/k=x
but I got the wrong answer
I've tried a couple different ways but cant seem to quite get it. I keep coming out with a negative answer for some reason.
I've tried Fnet=ma
Fnet=0
Fs1-Ff-Fs2-Ff=0
Fs1-2Ff=Fs2
(kL1-2uK*mg)/k=x
but I got the wrong answer
-
You have to take into account the energy lost through friction.
In the initial position, Ep0=k*x0^2/2
In the final rest position, Ep=k*x^2/2
The mechanical work of the friction force: W=F*d = m*u*g*(x-x0)
Ep0=Ep+W
k*x0^2/2 = k*x^2/2 + m*u*g*(x-x0)
93.8*(-0.257)^2/2 = 93.8*x^2/2 + 2.69*0.477*9.8*(x+0.257)
If you solve:x=-0.257 or x=-0.011 (the block will stop before reaching the 0 position)
The displacement: s=x-x0 = -0.011-(-0.257) = 0.246
In the initial position, Ep0=k*x0^2/2
In the final rest position, Ep=k*x^2/2
The mechanical work of the friction force: W=F*d = m*u*g*(x-x0)
Ep0=Ep+W
k*x0^2/2 = k*x^2/2 + m*u*g*(x-x0)
93.8*(-0.257)^2/2 = 93.8*x^2/2 + 2.69*0.477*9.8*(x+0.257)
If you solve:x=-0.257 or x=-0.011 (the block will stop before reaching the 0 position)
The displacement: s=x-x0 = -0.011-(-0.257) = 0.246