A hockey puck with a weight of 0.50 lbs is sliding freely across a section of ve

Question

A hockey puck with a weight of 0.50 lbs is sliding freely across a section of very smooth (frictionless) horizontal ice. (a)When it is sliding freely, how does the upward force of the ice on the puck (the normal force) compare to the upward force when the puck is sitting permantently at rest: (1) The upward force is greater when the puck is sliding; (2) The upward force is less when it is sliding; (3) The upward force is the same is both situations? Please explain why.

(b) Calculate the upward force on the puck in both situtations.

(It may help to draw force diagrams for both situations. Remember that horizontal and vertical forces are independent of eachother. Convert lbs to Newtons)

Explanation / Answer

a) In both the cases Upward force is equal to the downward force i.e 2.2148 N

b) According to Newton's second lawF = ma
F = (0.226)(9.8) = 2.2148N
Then we use newton's 3rd law to solve. F1 + F2 = 0. Hence, F1 = -F2.

So the upward force (including direction) = -2.2148 N (assuming down is positive and up is negative)
*I converted lbs to kg*

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