- #1

Like Tony Stark

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- Homework Statement
- The picture shows a block of ##m=2 kg## which moves along a cylinder with constant velocity ##1 \frac{m}{s}## relative to Earth due to the force applied in the photo. The coefficient of dynamic friction between the cylinder and the block is ##0.3## and ##0## between the cylinder and the Earth. Determine the distance with respect to the Earth that the block should move so that it moves ##0.05 m## with respect to the cylinder which was static at first.

- Relevant Equations
- Newton's equations

I called the block ##A## and ##B## the cylinder. For ##A##, in the non inertial system I wrote

##x) Fr_d -F=0##

##y) R-W_1=0##

Where ##Fr_d## is the kinetic friction and ##R## the normal force on ##A##

Then, I know that:

##a_{B/A}=a_B-a_A##

##0=a_B -a_A##

##a_A=a_B##

##v_{B/A}=v_B-v_A##

##v_{B/A}=v_B-(-1;0)##

But if the velocity of ##A## is constant, then it has no acceleration and so ##a_B=0## and then ##x_A=-t## so I just have to replace with ##-0.05##

##x) Fr_d -F=0##

##y) R-W_1=0##

Where ##Fr_d## is the kinetic friction and ##R## the normal force on ##A##

Then, I know that:

##a_{B/A}=a_B-a_A##

##0=a_B -a_A##

##a_A=a_B##

##v_{B/A}=v_B-v_A##

##v_{B/A}=v_B-(-1;0)##

But if the velocity of ##A## is constant, then it has no acceleration and so ##a_B=0## and then ##x_A=-t## so I just have to replace with ##-0.05##