Please show steps with formulas used so I will be able to repeat it on my own in
ID: 2007835 • Letter: P
Question
Please show steps with formulas used so I will be able to repeat it on my own in the future.
A 0.500 kg magnetic iron block sits on a level steel plate. The coefficient of static and kinetic friction between the block and the plane are 0.500 and 0.300, respectively. A horizontal force of magnitude 5.00 N is applied to the block, attempting to slide the block across the sheet. What must be the minumum magnetic force attracting the block to the sheet in order to keep it from sliding? Round to 2 significant figures.
_________N
Explanation / Answer
Since F=F, the normal force multiplied by the static coefficient of friction (coefficient while not moving) must equal the force applied to the block so they balance out at zero.
The static coefficient is .500, there fore the normal force must be double the force pushing the block, 10 newtons.
Since F=MA, the normal force the block is applying due to gravity is .500 kg x 9.81 m/s/s = 4.91 Newtons. The magnetic force needed is 10 - 4.91 = 5.09 Newtons.
Looking at it the other way. 5.09 N + 4.91 N = 10 N, parallel to the plate F=F, 10 N x .500 = 5 N, which is equal to the force horizontally. Since the forces equal zero the block is not moving.