Answer:
2.37 m/s
Explanation:
From the question
W = W'-Wf......................... Equation 1
Where W = net work done by Susan, W' = Work done by Susan, Wf = Work done against friction
W = FdcosФ-[d(mgμ-FsinФ)]................... Equation 2
Where F = the force applied by Susan, d = distance, Φ = angle of the force to the horizontal, m = mass, μ = coefficient of friction, g = acceleration due to gravity.
Given: F = 30 N, d = 3 m, m = 10 kg, μ = 0.2, g = 9.8 m/s², Ф = 30°
Substitute into equation 2
W = 30(3)(cos30°)-0.6[(9.8)(10)-30sin30°]
W = 77.94-49.8
W = 28.14 J.
But,
W = 1/2mv²........................ Equation 3
Where v = Paul's speed
make v the subject of the equation
v = √(2W/m).................. Equation 3
Given: W =28.14 J, m = 10 kg.
Substitute into equation 3
v = √(2×28.14/10)
v = √(56.28/10)
v =√5.628
v = 2.37 m/s
