Kaplunovsky - hw 07 - homework 07 Due: Mar 7 2007, 4:00 am...

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homework 07 – – Due: Mar 7 2007, 4:00 am 1 Question 1, chap 7, sect 1. part 1 of 1 6 points Stan does 151 J of work lifting himself 0 . 1 m. The acceleration of gravity is 9 . 8 m / s 2 . What is Stan’s mass? Correct answer: 154 . 082 kg (tolerance ± 1 %). Explanation: Let : w = 151 J , h = 0 . 1 m , and g = 9 . 8 m / s 2 . The work done is W = mg h m = W g h = 151 J (9 . 8 m / s 2 ) (0 . 1 m) = 154 . 082 kg . Question 2, chap 8, sect 2. part 1 of 1 8 points A bullet with a mass of 2 . 66 g and a speed of 673 m / s penetrates a tree horizontally to a depth of 4 . 08 cm. Assume that a constant frictional force stops the bullet. Hint: Try energy considerations. Calculate the magnitude of this frictional force. Correct answer: 14764 . 6 N (tolerance ± 1 %). Explanation: Let : m = 2 . 66 g , v = 673 m / s , and d = 4 . 08 cm . We can use conservation of energy to relate the initial kinetic energy of the bullet to the work done by the frictional force. 1 2 mv 2 = f · s Solving for the frictional force, f , f = mv 2 2 d = (0 . 00266 kg) (673 m / s) 2 2 (0 . 0408 m) = 14764 . 6 N . Question 3, chap 7, sect 3. part 1 of 1 8 points A 1240 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5 . 93 m before contacting the beam, and it drives the beam 6 . 38 cm into the ground before coming to rest. The acceleration of gravity is 9 . 8 m / s 2 . Using the work-energy theorem, calculate the magnitude of the average force the beam exerts on the pile driver while the pile driver is brought to rest. Correct answer: 1 . 14164 × 10 6 N (tolerance ± 1 %). Explanation: The work energy theorem tells us that the change in potential energy of the falling pile driver equals the work done on the I beam. We can express this relation as mg h = F · s, where h is the distance the pile driver falls, i.e. , h = (5 . 93 m) + (0 . 0638 m). Solving for the average force, F , we obtain F = mg h s = (1240 kg)(9 . 8 m / s 2 )(5 . 93 m + 0 . 0638 m) 0 . 0638 m = 1 . 14164 × 10 6 N . Question 4, chap 3, sect 4. part 1 of 1 8 points
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homework 07 – – Due: Mar 7 2007, 4:00 am 2 Vector v A has components A x = 2 . 93 , A y = 6 . 32 , A z = 1 . 06 while vector v B has components B x = 3 . 84 , B y = 6 . 8 , B z = 6 . 66 . What is the angle θ AB between these vectors? (Answer between 0 and 180 .) Correct answer: 130 . 708 (tolerance ± 1 %). Explanation: Consider two formul± for the scalar prod- uct v A · v B of two vectors: v A · v B = A x B x + A y B y + A z B z (1) in terms of the two vectors’ components, and also v A · v B = | v A || v B | cos θ AB (2) in term of their magnitudes and the angle be- tween them. Given the data, we immediately calculate | v A | = r A 2 x + A 2 y + A 2 z = 7 . 04634 , (3) | v B | = r B 2 x + B 2 y + B 2 z = 10 . 2636 , (4) and using eq. (1), v A · v B = 47 . 1676 . (5) Hence, according to eq. (2), cos θ AB = v A · v B | v A || v B | = 0 . 6522 (6) and therefore θ AB = arccos(
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This note was uploaded on 01/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Kaplunovsky - hw 07 - homework 07 Due: Mar 7 2007, 4:00 am...

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