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Kaplunovsky - hw 08

# Kaplunovsky - hw 08 - homework 08 Due 4:00 am Explanation...

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homework 08 – – Due: Mar 22 2007, 4:00 am 1 Question 1, chap 7, sect 2. part 1 of 2 5 points The force required to stretch a Hooke’s-law spring varies from 0 N to 22 N as we stretch the spring by moving one end 6 . 58 cm from its unstressed position. Find the force constant of the spring. Correct answer: 334 . 347 N / m (tolerance ± 1 %). Explanation: At the extended position, we can write Hooke’s law, ( F = k x ) solved for k . k = F x = 22 N 0 . 0658 m = 334 . 347 N / m Question 2, chap 7, sect 2. part 2 of 2 5 points Find the work done in stretching the spring. Correct answer: 0 . 7238 J (tolerance ± 1 %). Explanation: Having determined k , we can determine the work done in stretching the spring. This is just the energy stored in the spring, W = 1 2 k x 2 = 1 2 (334 . 347 N / m) (0 . 0658 m) 2 = 0 . 7238 J Question 3, chap 8, sect 1. part 1 of 1 8 points A toy gun is powered by a spring with k = 359 N / m; in the ”charged” position, the spring is compressed by 2 cm. The gun is loaded with a 15 g ball, the spring is compre- seed, the gun is aimed vertically up, and then ihe spring is released and the ball shoots out. How high with the ball rise above the point where the spring becomes un-compressed? Take g = 9 . 8 m / s 2 and neglect the air resis- tance and other frictional forces. Correct answer: 0 . 468435 m (tolerance ± 1 %). Explanation: Let : m = 15 g = 0 . 015 kg , x = 2 cm = 0 . 02 m , k = 359 N / m , and g = 9 . 8 m / s 2 . Applying conservation of energy for the mo- tion, U 0 + K s = U f m g x + 1 2 k x 2 = m g h Thus h = k x 2 2 m g x = (359 N / m) (0 . 02 m) 2 2 (0 . 015 kg) (9 . 8 m / s 2 ) 0 . 02 m = 0 . 468435 m . Question 4, chap 8, sect 1. part 1 of 2 10 points The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The block of mass m 1 lies on a rough horizontal surface with a constant coefficient of kinetic friction μ . This block is connected to a spring with spring constant k . The second block has a mass m 2 . The system is released from rest when the spring is unstretched, and m 2 falls a distance h before it reaches the lowest point. Note: When m 2 is at the lowest point, its velocity is zero. m 1 m 2 k m 1 m 2 h h μ Consider the moment when m 2 has de- scended by a distance s , where s is less than h .

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homework 08 – – Due: Mar 22 2007, 4:00 am 2 At this moment the sum of the kinetic energy for the two blocks K is given by 1. K = m 2 g s 1 2 k s 2 μ ( m 1 + m 2 ) g s . 2. K = m 2 g s + 1 2 k s 2 + μ ( m 1 + m 2 ) g s . 3. K = ( m 1 + m 2 ) g s + 1 2 k s 2 + μ m 1 g s . 4. K = m 2 g s 1 2 k s 2 μ m 1 g s . correct 5. K = ( m 1 + m 2 ) g s 1 2 k s 2 μ m 1 g s . 6. K = ( m 1 + m 2 ) g s 1 2 k s 2 + μ m 1 g s . 7. K = m 2 g s + 1 2 k s 2 + μ m 1 g s . 8. K = ( m 1 + m 2 ) g s + 1 2 k s 2 μ m 1 g s . Explanation: Basic Concepts: Work-Energy Theorem Spring Potential Energy Frictional Force according to the Work- Energy Theorem Solution: W ext A B = ( K B K A ) + ( U g B U g A ) + ( U sp B U sp A ) + W dis A B For the present case, the external work W ext A B = 0, A corresponds to the initial state and B the state where m 2 has descended by a distance s . The sum of the kinetic energy of m 1 plus that of m 2 at B is given by K = K B = ( U g A U g B ) + ( U sp A
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