Answer:
(a) <em>3 J and 4.899 m/s</em>
<em>(b) -3 J.</em>
<em>(c) </em>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.
<em>Note: Work done by the shuffleboard player = Kinetic Energy of the puck when the force is removed, assuming no energy is lost to friction.</em>
<em>Therefore,</em>
<em>Ek = 1/2mv² = F×d</em>
<em>Ek = F×d ............................................. Equation 3</em>
<em>Given: F = 6 N, d = 0.5 m.</em>
<em>Substituting these values into equation 3</em>
<em>Ek = 6×0.5</em>
<em>Ek = 3 J.</em>
<em>Thus the kinetic Energy = 3 J.</em>
Also,
Ek = 1/2mv²
making v the subject of the equation
v = √(2Ek/m)....................... Equation 4
<em>Given: m = 0.25, Ek = 3 J</em>
<em>Substituting into equation 4</em>
<em>v = √(2×3/0.25)</em>
<em>v = √24</em>
<em>v = 4.899 m/s</em>
<em>Thus the speed of the puck = 4.899 m/s</em>
<em>(b) </em><em>The work required to bring the puck to rest = -(the Kinetic Energy of the puck.)</em>
<em>Wₙ = 1/2mv²</em>
<em>Where Wₙ = work required to bring the puck to rest.</em>
<em>Where m = 0.25 kg, v = 4.899 m/s²</em>
<em>Wₙ = -1/2(0.25)(4.899)²</em>
<em>Wₙ = -1/2(0.25)(24)</em>
<em>Wₙ = -0.25(12)</em>
<em>Wₙ = -3 J</em>
<em>Thus the work required to bring the puck to rest = 3 J.</em>
(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.
