A child is pulling a wagon down the sidewalk. For 9.5 m the wagon stays on the sidewalk and the child pulls with a horizontal force of 23 N. Then one wheel of the wagon goes off on the grass so the child has to pull with a force of 40 N at an angle of 10 degrees to the side for the next 4.5 m. Finally the wagon gets back on the sidewalk so the child makes the rest of the trip, 15.0 m, with a force of 25 N. How much total work did the child do on the wagon?

W = 770.8 J

Explanation:

In order to find the total work done by the child, we need to find the work done during all three intervals. The formula for work done is as follows:

W = F d Cosθ

where,

W = Work done

F = Force Applied

d = displacement covered

θ = Angle between force and displacement

FOR 1ST INTERVAL

W₁ = F₁ d₁ Cosθ₁

where,

W₁ = Work done = ?

F₁ = Force Applied = 23 N

d₁ = displacement covered = 9.5 m

θ₁ = Angle between force and displacement = 0° (Parallel)

Therefore,

W₁ = (23 N)(9.5 m)(Cos 0°)

W₁ = 218.5 J

FOR 2ND INTERVAL

W₂ = F₂ d₂ Cosθ₂

where,

W₂ = Work done = ?

F₂ = Force Applied = 40 N

d₂ = displacement covered = 4.5 m

θ₂ = Angle between force and displacement = 10° (Parallel)

Therefore,

W₂ = (40 N)(4.5 m)(Cos 10°)

W₂ = 177.3 J

FOR 3RD INTERVAL

W₃ = F₃ d₃ Cosθ₃

where,

W₃ = Work done = ?

F₃ = Force Applied = 25 N

d₃ = displacement covered = 15 m

θ₃ = Angle between force and displacement = 0° (Parallel)

Therefore,

W₃ = (25 N)(15 m)(Cos 0°)

W₃ = 375 J

TOTAL WORK:

W = W₁ + W₂ + W₃

W = 218.5 J + 177.3 J + 375 J

W = 770.8 J

volts times amps 6 x 0.5 3w

area of focus