Work Problem and Solutions

 Problem #1

A floating ice block is pushed through a displacement of d = (15 m)i − (12 m)j along a straight embankment by rushing water, which exerts a force F = (210 N)i − (150 N)j on the block. How much work does the force do on the block during the displacement?

Answer:
Here we have the simple case of a straight–line displacement d and a constant force F. Then the work done by the force is W = F · d. We are given all the components, so we can compute the dot product using the components of F and d:

W = F · d = Fxdx + Fydy = (210 N)((15 m) + (−150 N)(−12 m) = 4950 J

Problem #2
A particle is subject to a force F_x that varies with position as in Fig. 1. Find the work done by the force on the body as it moves (a) from x = 0 to x = 5.0m, (b) from x = 5.0m to x = 10m and (c) from x = 10m to x = 15m. (d) What is the total work done by the force over the distance x = 0 to x = 15m?

Fig.1

Answer:
(a) Here the force is not the same all through the object’s motion, so we can’t use the simple formula W = F_xx. We must use the more general expression for the work done when a particle moves along a straight line,

W = F_x.Δx (Work = area of graph F vs. Displacement)

Of course, this is just the “area under the curve” of F_x vs. x from x_i to x_f . In part (a) we want this “area” evaluated from x = 0 to x = 5.0 m. From the figure, we see that this is just half of a rectangle of base 5.0m and height 3.0 N. So the work done is

W = ½ (3.0 N)(5.0m – 0) = 7.5J (area of a triangle)

(Of course, when we evaluate the “area”, we just keep the units which go along with the
base and the height; here they were meters and newtons, the product of which is a joule.)
So the work done by the force for this displacement is 7.5 J.

(b) The region under the curve from x = 5.0m to x = 10.0m is a full rectangle of base 5.0m and height 3.0 N. The work done for this movement of the particle is

W = (3.0 N)(5.0 m) = 15. J

(c) For the movement from x = 10.0 m to x = 15.0 m the region under the curve is a half rectangle of base 5.0 m and height 3.0 N. The work done is

W = ½ (3.0 N)(5.0m) = 7.5 J

(d) The total work done over the distance x = 0 to x = 15.0 m is the sum of the three
separate “areas”,

W total = 7.5 J +15 J + 7.5 J = 30 J

Problem #3
A 2 kg beam is driven by the force F = 40 N as far as 2 m along the inclined plane with slope 37⁰  in Fig. 2a. If the surface is rough and the friction force between the inclined plane and the beam is 10 N. Find the work done by (a) force F, (b) normal force, (c) gravity, (d) friction force, and (e) total work done by the force’s!

Fig.2

Answer:
The styles that work on the beam are shown as shown in Fig.2b!

(a) Work by force F in Fig.3(a).

       W_F = F.∆s cos θ = (40 N) (2 m) cos 00
       W_F = 80 J

Fig.3

 (b) Work by normal Force, in Fig.3(b).

       N = mg cos 370 = 24 N
       W_N = N.∆s cos θ = (24 N) (2 m) cos 900 = 0 J

(c) Work by gravity, in Fig 3(c).

       W = F.∆s cos θ
       W_g = (w)(∆s) cos θ = (20 N) (2 m) cos 1270 = - 24 J

(d) Work by friction, in Fig.3(d).

       W = F.∆s cos θ
       W_f = (f)(∆s) cos 180⁰ = (20 N)(2 m) cos 180⁰ = - 40 J

(e) total work done by the force’s

      W_total = W_F + W_f + W_g + W_N = 80 J + (-40 J) + (-24 J) + 0 = 16 J 

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