Problem#1
The force acting on a particle varies as in Figure 1. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x = 10.0 m, and (c) from x = 0 to x = 10.0 m.
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| Fig.1 |
Answer:
W = ∫F.dx = area under curve from xi to xf, then
(a) the work done by the force on the particle as it moves from x = 0 to x = 8.00 m,
W = area of triangle ABC = ½ AC x altitude
W0→8m = ½ (8.00 m)(6.00 N) = 24.0 J
(b) the work done by the force on the particle as it moves from x = 8.00 to x = 10.0 m,
W = area of triangle CDE = ½ CE x altitude
W8→10m = ½ (2.00 m)(–3.00 N) = –3.00 J
(c) the work done by the force on the particle as it moves from x = 8.00 to x = 10.0 m,
W0→10m = 24.0 J + (–3.00 J) = 21.0 J
Problem#2
The force acting on a particle is Fx = (8x – 16) N, where x is in meters. (a) Make a plot of this force versus x from x = 0 to x = 3.00 m. (b) From your graph, find the net work done by this force on the particle as it moves from x = 0 to x = 3.00 m.
Answer:
(a) The force acting on a particle is Fx = (8x – 16) N, then a plot of this force versus x from x = 0 to x = 3.00 m.
(b) find the net work done by this force on the particle as it moves from x = 0 to x = 3.00 m is
Wnet = ½ (2.00 m)(–16.0 N) + ½ (1.00 m)(8.00 N) = –12.0 J
Problem#3
A particle is subject to a force Fx that varies with position as in Figure 3. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) What is the total work done by he force over the distance x = 0 to x = 15.0 m?
Answer:
W = ∫Fdx and W equals the area under the Force-Displacement curve
(a) For the region 0 ≤ x ≤ 5.00 m,
W = ½ (3.00 N)(5.00m) = 7.50 J
(b) For the region 5.00 m ≤ x ≤ 10.0 m,
W = (3.00 N)(5.00m) = 15.0 J
(c) For the region 10.0m ≤ x ≤ 15.0 m,
W = ½ (3.00 N)(5.00m) = 7.50 J
(a) For the region 0 ≤ x ≤ 15.0 m,
W = 7.50 J + 15.0 J + 7.50 J = 30.0 J
Problem#4
A force F = (4xi + 3yj) N acts on an object as the object moves in the x direction from the origin to
x = 5.00 m. Find the work W = ∫F.dr done on the object by the force.
Answer:
W = ∫if F.dr
With F = (4xi + 3yj), then
W = ∫05 m(4xi + 3yj)dx
= 2x2│05m
W = 50.0J
Problem#5
When a 4.00-kg object is hung vertically on a certain light spring that obeys Hooke’s law, the spring stretches 2.50 cm. If the 4.00-kg object is removed, (a) how far will the spring stretch if a 1.50-kg block is hung on it, and (b) how much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?
Answer:
spring constant is given by k = mg/y = (4.00 kg)(9.80 m/s2)/(0.025 m) = 1.57 x 103 N/m, then
(a) For 1.50 kg mass
y = mg/k = (1.50 kg)(9.80 m/s2)/(1.57 x 103 N/m) = 9.38 x 10-3 m = 9.38 mm
(b) work must an external agent do to stretch the same spring 4.00 cm from its unstretched position, given by
W = ½ ky2
W = ½ (1.57 x 103 N/m)(4.00 x 10-2 m)2 = 1.25 J
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