Problem#1
A 200-g particle is released from rest at point A along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. 1). Calculate (a) the gravitational potential energy of the particle–Earth system when the particle is at point A relative to point B, (b) the kinetic energy of the particle at point B, (c) its speed at point B, and (d) its kinetic energy and the potential energy when the particle is at point C.
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| Fig.1 |
Answer:
(a) the gravitational potential energy of the particle–Earth system when the particle is at point A relative to point B,
UA = mgR = (0.200 kg)(9.80 m/s2)(0.300 m) = 0.588 J
(b) the kinetic energy of the particle at point B,
KA + UA = KB + UB
KB = KA + UA – UB = mgR = 0.588 J
(c) its speed at point B is
KB = ½ mvB2
0.588 J = ½ (0.200 kg)vB2
vB = 2.42 m/s
(d) its kinetic energy and the potential energy when the particle is at point C,
UC = mghC = (0.200 kg)(9.80 m/s2)(0.200 m) = 0.392 J
KC = KA + UA – UC = mg(hA – hC)
KC = (0.200 kg)(9.80 m/s2)(0.300 m – 0.200 m) = 0.196 J
Problem#2
What If? The particle described in Problem 1 (Fig.1) is released from rest at A, and the surface of the bowl is rough. The speed of the particle at A is 1.50 m/s. (a) What is its kinetic energy at B? (b) How much mechanical energy is transformed into internal energy as the particle moves from A to B? (c) Is it possible to determine the coefficient of friction from these results in any simple manner? Explain.
Answer:
(a) kinetic energy at B is
KB = ½ mvB2 = ½ (0.200 kg)(1.50 m/s)2 = 0.225 J
(b) ∆Emech = ∆K + ∆U = KB – KA + UB – UA
∆Emech = KB – 0 + mg(hB – hA)
∆Emech = 0.225 J + (0.200kg)(9.80 m/s2)(0 – 0.300 m)
∆Emech = –0.363 J
(c) It’s possible to find an effective coefficient of friction, but not the actual value of µ since n
and f vary with position.
Problem#3
A 2.00-kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m (Fig. 2). The pulley is frictionless. The block is released from rest when the spring is unstretched. The block moves 20.0 cm down the incline before coming to rest. Find the coefficient of kinetic friction between block and incline.
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| Fig.1 |
Answer:
The gain in internal energy due to friction represents a loss in mechanical energy that must be equal
to the change in the kinetic energy plus the change in the potential energy.
Therefore,
∆Emech = ∆K + ∆U
–fx = ∆K + ½ kx2 + 0 – mgx sin37.00
–µkmgxcos37.00 = ∆K + ½ kx2 + 0 – mgx sin37.00
–µkmgxcos37.00 = ∆K + ½ kx2 – mgx sin37.00
But, ∆K = 0, because vi = vf = 0, then
–µkmgxcos37.00 = ½ kx2 – mgx sin37.00
–µk(2.00 kg)(9.80m/s2)(0.200 m)cos37.00 = ½ (100 N/m)(0.200 m)2 – (2.00 kg)(9.80 m/s2)(0.200 m)sin37.00
–µk(3.13) = –0.359
µk = 0.115
Note that in the above we had a gain in elastic potential energy for the spring and a loss in gravitational potential energy.
Problem#4
Review problem. Suppose the incline is frictionless for the system described in Problem 3 (Fig. 2). The block is released from rest with the spring initially unstretched. (a) How far does it move down the incline before coming to rest? (b) What is its acceleration at its lowest point? Is the acceleration constant? (c) Describe the energy transformations that occur during the descent.
Answer:
(a) Since no nonconservative work is done, ∆E = 0. Also ∆K = 0, therefore Uf = Ui, with
Ui = (mgsinθ)x and
Uf = ½ kx2
So that
½ kx2 = mgxsinθ
½ (100 N/m)x = (2.00 kg)(9.80 m/s2)sin37.00
x = 0.236 m
(b) ∑F = ma. Only gravity and the spring force act on the block, so
–kx + mg sinθ = ma
–(100 N/m)(0.236 m) + (2.00 kg)(9.80 m/s2) sin37.00 = (2.00 kg)a
a = –5.90 m/s2. The negative sign indicates a is up the incline. The acceleration depends on position.
(c) U(gravity) decreases monotonically as the height decreases.
U(spring) increases monotonically as the spring is stretched.
K initially increases, but then goes back to zero
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