1. A bead of mass 1.10 kg (yes, that\'s a big bead) is sliding along a frictionl

Question

1. A bead of mass 1.10 kg (yes, that's a big bead) is sliding along a frictionless wire. Consider the lowest line (Point H) to be the reference level for the sake of calculating potential energy. Each division is 3.40 m tall. What is the potential energy of the bead at Point K?

2. Suppose the bead is traveling at a speed of 11.5 m/s at Point K. What is the kinetic energy of the bead at Point K?

3. The total mechanical energy is defined as the sum of the kinetic and potential energy. What is the total mechanical energy of the bead at Point K?

4. As the bead slides along the frictionless wire, there is no loss of total mechanical energy. What is the total mechanical energy of the bead at Point L?

5. What is the potential energy of the bead at Point L?

6. What is the kinetic energy of the bead at Point L?

7. How fast is the bead traveling at Point L?

Explanation / Answer

1

hk = height of K above H = 3.40 m

Potential energy at K = mg hk = 1.10 x 9.8 x 3.40 = 36.65 J

2.

Vk = speed at K = 11.5 m/s

Kinetic energy at K = (0.5) m Vk2 = (0.5) (1.10) (11.5)2 = 72.74 J

.3.

Total mechanical energy = KE + PE = 36.65 + 72.74 = 109.4 J

4.

Using conservation of energy between point L and K

Total mechanical energy at L = Total mechanical energy at K = 109.4 J

5.

hL = height at L = 2 x 3.40 = 6.80 m

Potential energy at L = mg hL = 1.10 x 9.8 x 6.80 = 73.3 J

6.

PEL + KEL = Total mechanical energy at L = Total mechanical energy at K = 109.4 J

73.3 + KEL = 109.4

KEL = 36.1 J

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