Question

A shuffleboard player pushes a 0.25 kg puck, initially at rest, such that a constant horizontal force of 6 N acts on it through a distance of 0.5 m. (Neglect friction.)a. What are the kinetic energy and the speed of the puck when the force is removed?b. How much work would be required to bring the puck to rest?c. FOLLOW UP: Suppose the puck in this problem had twice the final speed. Would it then take twice as much work to stop the puck?

Answer 1

Answer:

(a) 3 J and 4.899 m/s

(b) -3 J.

(c) It will take four times as much as the work done to stop it when the final speed is increased twice.

Explanation:

(a) Work done by the shuffleboard player = Force × distance

W = F×d.................................. Equation 1

Where W = work done, F = Force, d = distance.

Kinetic Energy (Ek) of the = 1/2mv²...................... Equation 2

Where m = mass of the puck, v = velocity of the puck.

Note: Work done by the shuffleboard player = Kinetic Energy of the puck when the force is removed, assuming no energy is lost to friction.

Therefore,

Ek = 1/2mv² = F×d

Ek = F×d ............................................. Equation 3

Given: F = 6 N, d = 0.5 m.

Substituting these values into equation 3

Ek = 6×0.5

Ek = 3 J.

Thus the kinetic Energy = 3 J.

Also,

Ek = 1/2mv²

making v the subject of the equation

v = √(2Ek/m)....................... Equation 4

Given: m = 0.25, Ek = 3 J

Substituting into equation 4

v = √(2×3/0.25)

v = √24

v = 4.899 m/s

Thus the speed of the puck = 4.899 m/s

(b) The work required to bring the puck to rest = -(the Kinetic Energy of the puck.)

Wₙ = 1/2mv²

Where Wₙ = work required to bring the puck to rest.

Where m = 0.25 kg, v = 4.899 m/s²

Wₙ = -1/2(0.25)(4.899)²

Wₙ = -1/2(0.25)(24)

Wₙ = -0.25(12)

Wₙ = -3 J

Thus the work required to bring the puck to rest = 3 J.

(c) Assuming the puck has twice the final speed

Work required to stop it.

Wₓ = 1/2mv²

Where Wₓ = work required to stop the puck when it has twice the final speed.

m = 0.25 kg, v = 9.798 m/s ( twice the final speed)

Wₓ = 1/2(0.25)(9.798)²

Wₓ = 1/2(0.25)(96)

Wₓ = 12 J.

Thus the puck will take four times as much as the work done to stop it when the final speed is increased twice.

Answer 2

(a) The kinetic energy of the puck is 3 J and speed is 4.899 m/s

(b) 3 J  work against the motion would be required to bring the puck to rest

(c) It will take four times as much as the work done to stop it when the final speed is increased twice.

Given that the puck m = 0.25 kg, horizontal force F = 6N, distance d = 0.5m

(a) Work done by the shuffleboard player = Force × distance

     W = F×d

where W = work done, F = Force, d = distance.

Kinetic Energy KE = 1/2mv²

where m = mass of the puck, v = velocity of the puck.

work-energy theorem:

From work-energy theorem:

change in KE = work done

KE = F×d = 6 × 0.5J

KE = 3J will be the kinetic energy of the puck

Also, from KE = W, we get:

1/2mv² = F×d

v² = 2(F×d)/m

v² = (2×6×0.5/0.25)

v² = 24

v = 4.899 m/s

Thus the speed of the puck = 4.899 m/s

(b) The work required to bring the puck to rest = the Kinetic Energy of the puck

W' = 1/2mv²

W'  = 1/2(0.25)(4.899)²

W'  = 1/2(0.25)(24)

W'  = 0.25(12)

W'  = 3 J

Thus the work required to bring the puck to rest = 3 J.

(c) Assuming the puck has twice the final speed

Work required to stop it is the same amount of energy against the kinetic energy of the puck

W" = 1/2mv²

W" = 1/2(0.25)(9.798)²

W" = 1/2(0.25)(96)

W" = 12 J

The puck will take four times as much as the work done to stop it when the final speed is increased twice.

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