Problem#1
A child’s pogo stick (Fig.1) stores energy in a spring with a force constant of 2.50 x 104 N/m. At position A (xA = –0.100 m), the spring compression is a maximum and the child is momentarily at rest. At position B (xB = 0), the spring is relaxed and the child is moving upward. At position C, the child is again momentarily at rest at the top of the jump. The combined mass of child and pogo stick is 25.0 kg. (a) Calculate the total energy of the child–stick–Earth system if both gravitational and elastic potential energies are zero for x = 0. (b) Determine xC. (c) Calculate the speed of the child at x = 0. (d) Determine the value of x for which the kinetic energy of the system is a maximum. (e) Calculate the child’s maximum upward speed. |
| Fig.1 |
Answer:
Given k = 2.50 x 104 N/m, m = 2.50 kg, xA = –0.100 m, then
(a) the total energy of the child–stick–Earth system if both gravitational and elastic potential energies are zero for x = 0,
Emech = KA + UgA + UsA
Emech = 0 + mgxA + ½kxA2 = (2.50 kg)(9.80 m/s2)(–0.100 m) + ½ (2.50 x 104 N/m)(–0.100 m)2
Emech = 100 J
(b) Since only conservative forces are involved, the total energy of the child-pogo-stick-Earth
system at point C is the same as that at point A.
KC + UgC + UsC = KA + UgA + Ugs
0 + mgxC + ½ kxC2 = 0 + mgxA + ½ kxA2
(25.0 kg)(9.80 m/s2)xC + 0= (2.50 kg)(9.80 m/s2)(–0.100 m) + ½ (2.50 x 104 N/m)(–0.100 m)2
245xC = –24.5 + 125
xC = 0.410 m
(c) the speed of the child at x = 0, we use
KB + UgB + UsB = KA + UgA + Ugs
½ mvB2 + mgxB + ½ kxB2 = 0 + mgxA + ½ kxA2
½ (25.0 kg)vB2 + 0 + 0 = 0 + (2.50 kg)(9.80 m/s2)(–0.100 m) + ½ (2.50 x 104 N/m)(–0.100 m)2
½ (25.0kg)vB2 = –24.5 + 125
12.5vB2 = –24.5 + 125
vB = 2.84 m/s
(d) K and v are at a maximum when a = ∑F/m (i.e., when the magnitude of the upward
spring force equals the magnitude of the downward gravitational force).
This occurs at x < 0 where
k│x│ = mg
x = (25.0 kg)(9.80 m/s2)/(2.50 x 104 N/m) = 9.80 mm
Thus, K = Kmax at x = –9.80 mm
(e) Suppose the point x = –9.80 m is the point Z, then
Kmax = KA + (UgA – UgZ) + (UsA – UsZ)
½ mvmax2 = 0 + mg(hA - hZ) + ½ k(xA2 – xZ2)
½ (25.0 kg)vmax2 = 0 + (25.0 kg)(9.80 m/s2)[–0.100 m – (–9.80 x 10-3 m)] + ½ (2.50 x 104 N/m)[(–0.100 m)2 – (–9.80 x 10-3 m)2]
(12.5 kg)vmax2 = –22.1 J + 124 J
vmax = 2.85 m/s
Problem#2
A 10.0-kg block is released from point A in Figure 2. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2 250 N/m, and compresses the spring 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C.
 |
| Fig.2 |
Given: m = 10.0 kg, dBC = 6.00 m, k = 2250 N/m, x = 0.300 m, we use
∆Emech = –f∆x
Ef – Ei = –µmgdBC
½kx2 – mgh = –µmgdBC
½ (2250 N/m)(0.300 m)2 – (10.0 kg)(9.80 m/s2)(3.00 m) = –µ(10.0 kg)(9.80 m/s2)(6.00 m)
101.25 J – 294 J = –µ(588 J)
µ = 0.328
Problem#3
The potential energy function for a system is given by U(x) = –x3 + 2x2 + 3x. (a) Determine the force Fx as a function of x. (b) For what values of x is the force equal to zero? (c) Plot U(x) versus x and Fx versus x, and indicate points of stable and unstable equilibrium.
Answer:
(a) the force Fx as a function of x is
F = –dU/dx
F = –d(–x3 + 2x2 + 3x)/dx
F = 3x2 – 4x + 3
(b) values of x is the force equal to zero we get
0 = 3x2 – 4x + 3
x = {4 ± √[(–4)2 + 4(3)(3)]}/6
x = –0.535 m or x = 1.87 m
(c) Plot U(x) versus x and Fx versus x, and indicate points of stable and unstable equilibrium.
The stable point is at x = -0.535 point of minimum U(x).
The unstable point is at x = 1.87 maximum in U(x).
Problem#4
A 20.0-kg block is connected to a 30.0-kg block by a string that passes over a light frictionless pulley. The 30.0-kg block is connected to a spring that has negligible mass and a force constant of 250 N/m, as shown in Figure 4. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0-kg block is pulled 20.0 cm down the incline (so that the 30.0-kg
block is 40.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0-kg block is 20.0 cm above the floor (that is, when the spring is unstretched).
 |
| Fig.4 |
We use
Ki + Ui = Kf + Uf
0 + mgx + ½ kx2 = ½ mv2 + mgxsin40.00
(30.0 kg)(9.80 m/s2)(0.200 m) + ½ (250 N/m)(0.200 m)2 = ½ (50.0 kg)v2 + (20.0 kg)(9.80 m/s2)(0.200 m) sin40.00
58.8 J + 5.00 J = (25.0 kg)v2 + 25.2 J
v = 1.24 m/s
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