Problem#1
A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. 1). The object has a speed of vi = 3.00 m/s when it makes contact with a light spring that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a
distance d. The object is then forced toward the left by the spring and continues to move in that direction beyond the spring’s unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring. Find (a) the distance of compression d, (b) the speed v at the
unstretched position when the object is moving to the left, and (c) the distance D where the object comes to rest.
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| Fig.1 |
Answer:
(a) Between the second and the third picture,
∆Emech = ∆K + ∆U
–µmgd = – ½ mvi2 + ½ kd2
–(0.250)(1.00 kg)(9.80 m/s2)d = – ½ (1.00 kg)(3.00 m/s)2 + ½ (50.0 N/m)d2
–2.45d = –4.50 + 25.0d2
25.0d2 + 2.45d – 4.50 = 0
d = {–2.45 ± [(–2.45)2 – 4(25.0)(–4.50)]1/2}N/[50.0 N/m]
d = {–2.45 ± 21.25}N/[50.0 N/m]
d = 0.378 m
(b) Between picture two and picture four,
∆Emech = ∆K = Kf – Ki
–f(2d) = ½ mv2 – ½ mvi2
–(2.45 N)(2)(0.378 m) = ½ (1.00 kg)v2 – ½ (1.00 kg)(3.00 m/s)2
–1.8522 J = 0.500kgv2 – 4.50 J
v = 2.30 m/s
(c) For the motion from picture two to picture five,
∆Emech = ∆K + ∆U
–f(2d +D) = – ½ mvi2
–(0.250)(1.00 kg)(9.80 m/s2)[2(0.378 m + D)]= – ½ (1.00 kg)(3.00 m/s)2
–(2.45 J)[2(0.378 m + D)]= –4.50 J
2(0.378 m) + D = 1.84 m
0.756 m + D = 1.84 m
D = 1.08 m
Problem#2
A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (Fig. 2). The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point B, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The speed of the block at the bottom of the track is vB = 12.0 m/s, and the block experi ences an average friction force of 7.00 N while sliding up the track. (a) What is x? (b) What speed do you predict for the block at the top of the track? (c) Does the block actually reach the top of the track, or does it fall off before reaching the top?
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| Fig.2 |
Answer:
(a) Initial compression of spring:
½ kx2 = ½ mv2
(450 N/m)(∆x)2 = (0.500 kg)(12.0 m/s)2
∆x = 0.400 m
(b) Speed of block at top of track:
∆Emech = ∆U + ∆K
–f∆x = mghT – mghB + ½ mvT2 – ½ mvB2
–f(πR) = m[ghT – ghB + ½vT2 – ½vB2]
–(7.00 N)π(1.00 m) = (0.500 kg)(9.80 m/s2)(2.00 m) – 0 + ½(0.500 kg)vT2 – ½(0.500 kg)(12.0 m/s)2
0.250vT2 = 4.21
vT = 4.10 m/s
(c) Does block fall off at or before top of track? Block falls if ac < g.
ac = v2/R = (4.10 m/s)2/(1.00 m) = 16.8 m/s2
Therefore ac > g and the block stays on the track.
Problem#3
A uniform chain of length 8.00 m initially lies stretched out on a horizontal table. (a) If the coefficient of static friction between chain and table is 0.600, show that the chain will begin to slide off the table if at least 3.00 m of it hangs over the edge of the table. (b) Determine the speed of the chain as all of it leaves the table, given that the coefficient of kinetic friction between the chain and the table is 0.400.
Answer:
Let λ represent the mass of each one meter of the chain and T represent the tension in the chain at the table edge.
(a) For the five meters on the table with motion impending,
∑Fy = 0
n – 5λg = 0
n = 5λg
Static friction is given by
fs = µsn = 0.600(5λg) = 3λg
∑Fx = 0
T – fs = 0, then T = fs = 3λg
The maximum value is barely enough to support the hanging segment according to
∑Fy = 0 (for the three meters )
T – 3λg = 0
T = 3λg
so it is at this point that the chain starts to slide.
(b) Let X represent the variable distance the chain has slipped since the start. Then length (5 – x) remains on the table, with now
∑Fy = 0
n – (5 – x)λg = 0
n = (5 – x)λg
then
fk = µkn = 0.400(5 – x)λg = 2λg – 0.4xλg
Consider energies of the chain-Earth system at the initial moment when the chain starts to slip, and a final moment when x = 5 , when the last link goes over the brink. Measure heights above the final position of the leading end of the chain. At the moment the final link slips off, the center of the chain is at yf = 4 meters.
Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at height 8 – 3/2 = 6.5 m. So, by using
∆Emech = ∆K + ∆U
–∫iffkdx = ½ mvf2 – 0 + mgyf – (m1gy1i + m2gy2i)
–∫05.00(2λg – 0.4xλg)dx = ½ (8λ)vf2 + (8λ)g(4) – (5λ)g(8) – (3λ)g(6.5)
–2g∫05.00dx + 0.4g∫05.00 xdx = 4vf2 – 27.5g
–2gx│05.00 + 0.2gx2│05.00 = 4vf2 – 27.5g
–2g(5.00 m) + 0.2g(5.00 m)2 = 4vf2 – 27.5g
4vf2 = 22.5(9.80 m/s2)
vf = 7.42 m/s
Problem#4
A child slides without friction from a height h along a curved water slide (Fig. 4). She is aunched from a height h/5 into the pool. Determine her maximum airborne height y in terms of h and θ.
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| Fig.4 |
Answer:
Launch speed is found from
mg(h – h/5) = ½ mv2
mg(4h/5) = ½ mv2
v = [8gh/5]1/2
with vy = vsinθ
The height y above the water (by conservation of energy for the child-Earth system) is found from
mgy = ½ mvy2 + mgh/5
mgy = ½ m(8gh/5)sin2θ + mgh/5
y = (h/5)(4 sin2θ + 1)
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