Problem#1
A 4.00-kg particle moves from the origin to position C, having coordinates x = 5.00 m and y = 5.00 m. One force on the particle is the gravitational force acting in the negative y direction (Fig. 1). Using Equation 7.3, calculate the work done by the gravitational force in going from O to C along (a) OAC. (b) OBC. (c) OC. Your results should all be identical. Why? |
| Fig.1 |
(a) the work done by the gravitational force in going from O to C along OAC
WOAC = W along OA + W along AC
WOAC = mg(OA)cos90.00 + mg(AC)
WOAC = 0 + (4.00 kg)(5.00 m)(9.80 m/s2)(–1) = –196 J
(b) the work done by the gravitational force in going from O to C along OBC
WOAC = W along OB + W along BC
WOAC = mg(OB)cos1800 + mg(BC)
WOAC = (4.00 kg)(5.00 m)(9.80 m/s2)(–1) + 0 = –196 J
(a) the work done by the gravitational force in going from O to C along OC
W along OC = mg(OC) cos1350
= mg(OA)cos90.00 + mg(AC)
= (4.00 kg)(9.80 m/s2)(5.00√2 m)(– ½√2) = –196 J
The results should all be the same, since gravitational forces are conservative.
Problem#2
(a) Suppose that a constant force acts on an object. The force does not vary with time, nor with the position or the velocity of the object. Start with the general definition for work done by a force
W = ∫ifF.dr
and show that the force is conservative. (b) As a special case, suppose that the force F = (3i + 4j) N acts on a particle that moves from O to C in Figure 1. Calculate the work done by F if the particle moves along each one of the three paths OAC, OBC, and OC. (Your three answers should be identical.)
Answer:
(a) W = ∫F.dr and if the force is constant, this can be written as
W = F. ∫dr = F(rf – ri), which depends only on end points, not path.
(b) W = ∫F.dr = ∫(3i + 4j)d(xi + yj)
W = ∫05.00 (3N)dx + ∫05.00 (4 N)dy
W = 15.0 J + 20.0 J = 35.0 J
The same calculation applies for all paths.
Problem#3
A force acting on a particle moving in the xy plane is given by F = (2yi + x2j), where x and y are in meters. The particle moves from the origin to a final position having coordinates x = 5.00 m and y = 5.00 m, as in Figure 1. Calculate the work done by F along (a) OAC, (b) OBC, (c) OC. (d) Is F conservative or nonconservative? Explain.
Answer:
(a) the work done by F along OAC
WOA = ∫F.dr = ∫(2yi + x2j)idx = ∫2ydx
and since along this path, y = 0, then WOA = 0.
WAC = ∫(2yi + x2j)jdy = ∫05.00 m x2dy = x2y│05.00 m
For, x = 5.00 m, then
WAC = (5.00 m)2(5.00 m) = 125 J
So that
WOAC = 0 + 125 J = 125 J
(b) the work done by F along OBC
since along this path, x = 0, WOB = 0
WBC = ∫(2yi + x2j)idx = ∫05.00 m 2ydx = 2yx│05.00 m
For, y = 5.00 m, then
WAC = 2(5.00 m)(5.00 m) = 50.0 J
So that
WOAC = 0 + 50.0 J = 50.0 J
(c) the work done by F along OC
WOC = ∫F.dr = ∫(2yi + x2j)d(xi + yj) = ∫(2ydx + x2dy)
Since x = y along OC
WOC = ∫05.00 m[2x + x2]dx = 66.7 J
(d) F is nonconservative since the work done is path dependent.
Problem#4
A particle of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure 2. Determine (a) the particle’s speed at points B and C and (b) the net work done by the gravitational force in moving the particle from A to C.
Answer:
(a) the particle’s speed at points B and C, we use
∑WAB = ∆KAB
Wg = KB - KA
–mg(hB – hA) = ½ mvB2 – ½ mvA2
–(9.80 m/s2)(3.20 m – 5.00 m) = ½ vB2 – 0
vB = 5.94 m/s
Similarly,
–mg(hC – hA) = ½ mvc2 – ½ mvA2
–(9.80 m/s2)(2.00 m – 5.00 m) = ½ vc2 – 0
vC = 7.67 m/s
(b) the net work done by the gravitational force in moving the particle from A to C is
WA→C = mg(∆h) = (5.00 kg)(9.80 m/s2)(3.00 m) = 147 J
Problem#5
A single constant force F = (3.00i + 5.00j) N acts on a 4.00-kg particle. (a) Calculate the work done by this force if theparticle moves from the origin to the point having the vector position r = (2.00i – 3.00j) m . Does this result depend on the path? Explain. (b) What is the speed of the particle at r if its speed at the origin is 4.00 m/s? (c) What is the change in the potential energy?
Answer:
(a) the work done by this force if theparticle moves from the origin to the point having the vector position r = (2i – 3j) m, given by
W = F.r = (3.00i + 5.00j)N.(2.00i – 3.00j)m
W = 6.00 – 15.0 J = –9.00 J
The result does not depend on the path since the force is conservative.
(b) the speed of the particle at r if its speed at the origin is 4.00 m/s is
W = ∆K
W = ½ mvf2 – ½ mvi2
–9.00 J = ½ (4.00 kg)v2 – ½ (4.00 kg)(4.00 m/s)2
23.0 J – 9.00 J = 2.00kgv2
v = 3.39 m/s
(c) the change in the potential energy given by
∆U = –W = 9.00 J
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