Problem#1
An archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. (a) What is the equivalent spring constant of the bow? (b) How much work does the archer do in pulling the bow?
Answer:
(a) Spring constant is given by F = kx, then
k = F/x = (230 N)/(0.400 m) = 575 N/m
(b) work does the archer do in pulling the bow is
W = Favgx = ½ (230 N)(0.400 m) = 46.0 J
Problem#2
Truck suspensions often have “helper springs” that engage at high loads. One such arrangement is a leaf spring with a helper coil spring mounted on the axle, as in Figure 1. The helper spring engages when the main leaf spring is compressed by distance y0, and then helps to support any additional load. Consider a leaf spring constant of 5.25 x 105 N/m, helper spring constant of
3.60 x 105 N/m, and y0 = 0.500 m. (a) What is the compression of the leaf spring for a load of 5.00 x 105 N? (b) How much work is done in compressing the springs?
Answer:
(a) the compression of the leaf spring for a load of 5.00 x 105 N, given by
Fapplied = kleafxl + khelperxh
5 x 105 N = (5.25 x 105 N/m)xl + (3.60 x 105 N/m)(xl – 0.500 m)
xl = (6.80 N)/(8.85 N/m)
xl = 0.768 m = 76.8 cm
(b) work is done in compressing the springs given by
W = ½ klxl2 + ½ khxh2
W = ½(5.25 x 105 N/m)(0.768 m)2 + ½(3.60 x 105 N/m)(0.268 m)2
W = 1.68 x 105 J
Problem#3
A 100-g bullet is fired from a rifle having a barrel 0.600 m long. Assuming the origin is placed where the bullet begins to move, the force (in newtons) exerted by the expanding gas on the bullet is 15,000 + 10,000x – 25,000x2, where x is in meters. (a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. (b) What If? If the barrel is 1.00 m long, how much work is done, and how does this value compare to the work calculated in (a)?
Answer:
(a) the work done by the gas on the bullet as the bullet travels the length of the barrel,
W = ∫F.dr
W = ∫00.6 m(15,000 + 10,000x – 25,000x2)dx
= 15,000x + 5,000x2 – (25,000/3)x3│00.6 m
= 9,000 J + 1,800 J – 1,800 J = 9.00 kJ
(b) Similarly,
W = ∫01.00 m(15,000 + 10,000x – 25,000x2)dx
= 15,000x + 5,000x2 – (25,000/3)x3│01.00 m
= 15,000 J + 5,000 J – 8,333 J = 11.7 kJ, larger by 29.6%
Problem#4
If it takes 4.00 J of work to stretch a Hooke’s-law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 10.0 cm.
Answer:
Given: W = 4.00 J, x = 10.0 cm, then
W = ½ kx2
4.00 J = ½ k(0.100 m)2
k = 800 N/m
and to stretch the spring to 0.200 m requires
ΔW = ½ (800 N/m)(0.200 m)2 – 4.00 J = 12.0 J
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