Three forces act on a hockey pick as it slides across the ice; the normal force N represents the force the ice exerts upward on the puck and is given by N= <0,0,100>, the gravitational force G represents the downward force of gravity on the puck and is given by G = <0,0,-100>, and the force due to friction F has a magnitude |F|=μ|N| and is in the direction opposite of whatever direction the puck is traveling, where μ is the coefficient of kinetic friction. Work (W) is done by any force A is given by W = A.s where s is the displacement vector associate with an object's motion. If an object moves away from (x1,y1,z1) to (x2,y2,z2), it's displacement can be represented by s = (x2-x1) xhat + (y2-y1) yhat = (z2-z1) zhat.

Consider a puck that moves from (1,1,0) to (3,3,0) on very slippery ice with a coefficient of kinetic friction of .15 and answer the following questions:
a. WHat is the displacement s of the puck? Write your answer in both component form and magnitude/direction.
b. What is the force of friction F acting on the puck? Write your answer in component form rounding to 1 decimal place.
c. How much work does friction do on the puck?
d. How much work does gravity do on the puck?
e. How much work does the normal force do on the puck?

To find the answers to the given questions, we'll follow the steps and formulas provided in the question.

a. The displacement s of the puck can be found using the formula:
s = (x2 - x1) x̂ + (y2 - y1) ȳ + (z2 - z1) ẑ

Given the initial position (x1,y1,z1) = (1,1,0) and final position (x2,y2,z2) = (3,3,0), we can substitute these values into the formula:

s = (3 - 1) x̂ + (3 - 1) ȳ + (0 - 0) ẑ
= 2x̂ + 2ȳ + 0ẑ

Therefore, the displacement s of the puck is (2, 2, 0) in component form.

The magnitude/direction of the displacement can be found using the formula:
|s| = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)

Substituting the values, we have:
|s| = √((3 - 1)^2 + (3 - 1)^2 + (0 - 0)^2)
= √(4 + 4 + 0)
= √8
= 2√2

Therefore, the magnitude of the displacement is 2√2, and the direction is along the vector (2, 2, 0).

b. The force of friction F acting on the puck can be calculated using the formula:
F = -μN, where N is the given normal force and μ is the coefficient of kinetic friction.

Given N = <0, 0, 100> and μ = 0.15, we can substitute these values into the formula:
F = -0.15 * <0, 0, 100>
= <0, 0, -15>

Therefore, the force of friction F acting on the puck is <0, 0, -15>.

c. The work done by friction can be found using the formula:
W = F · s, where F is the force of friction and s is the displacement vector.

Given F = <0, 0, -15> and s = (2, 2, 0), we can substitute these values into the formula:
W = <0, 0, -15> · (2, 2, 0)
= 0 + 0 + 0
= 0

Therefore, the work done by friction on the puck is 0.

d. The work done by gravity can be found using the formula:
W = G · s, where G is the gravitational force and s is the displacement vector.

Given G = <0, 0, -100> and s = (2, 2, 0), we can substitute these values into the formula:
W = <0, 0, -100> · (2, 2, 0)
= 0 + 0 + 0
= 0

Therefore, the work done by gravity on the puck is 0.

e. The work done by the normal force can be found using the formula:
W = N · s, where N is the normal force and s is the displacement vector.

Given N = <0, 0, 100> and s = (2, 2, 0), we can substitute these values into the formula:
W = <0, 0, 100> · (2, 2, 0)
= 0 + 0 + 0
= 0

Therefore, the work done by the normal force on the puck is 0.