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Unformatted text preview: le (txl65) – Homework 5 – Spurlock – (11881) This printout should have 5 questions. Multiplechoice questions may continue on the next column or page – ﬁnd all choices before answering. 001 (part 1 of 2) 10.0 points A force F = Fx ˆ + Fy ˆ acts on a particle that ı undergoes a displacement of s = sx ˆ + sy ˆ ı where Fx = 9 N, Fy = −4 N, sx = 3 m, and sy = 2 m. Find the work done by the force on the particle. Correct answer: 19 J. Explanation: The work is given by W = F · s = Fx sx + Fy sy = (9 N) (3 m) + (−4 N) (2 m) = 19 J . 002 (part 2 of 2) 10.0 points Find the angle between F and s. Correct answer: 57.6526◦. Explanation: Since W = F · s = F s cos θ W θ = cos−1 . F s F=
2 2 Fx + Fy = 1 003 10.0 points An applied force varies with position according to F = k1 xn − k2 , where n = 3, k1 = 3.5 N/m3 , and k2 = 34 N. How much work is done by this force on an object that moves from xi = 5.04 m to xf = 19 m? Correct answer: 112.992 kJ. Explanation: Basic Concepts: W = F · ds Solution: The work done by a varying force is
x2 W=
x1 F · ds. which adds up all the little F · s parts along the path, taking into account the changing force. Here all the motion is in the ˆ direction ı so ds = dx.
xf W= = F · dx
xi 19 m (3.5 N/m3 ) x3 − (34 N) · dx
5.04 m 19 m ( 3 . 5 N / m3 ) x 4 = − (34 N) x 4 5.04 m 3 . 5 N / m3 = (19 m)4 − (5.04 m)4 4 − (34 N) (19 m − 5.04 m) = 1.12992 × 105 J = 112.992 kJ . (9 N)2 + (−4 N)2 004 10.0 points A 83.9 g bullet is ﬁred from a riﬂe having a barrel 0.699 m long. Assuming the origin is placed where the bullet begins to move, the force exerted on the bullet by the expanding gas is F = a + b x − c x2 , where a = 23100 N, b = 12500 N/m, c = 24000 N/m2 , with x in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. = 9.84886 N and s = s2 + s2 = x y 3 m2 + 2 m2 so = 3.60555 m , θ = cos−1 19 J (9.84886 N) (3.60555 m) = 57.6526◦ . le (txl65) – Homework 5 – Spurlock – (11881)
28 m 2 Correct answer: 16468.4 J. Explanation: The work is found by integrating the force over the distance.
xf =
1.9 m (8.9 N/m3 ) x3 · dx
28 m = W=
xi F · ds ( 8 . 9 N / m3 ) x 4 4 1.9 m 3 8. 9 N /m (28 m)4 − (1.9 m)4 = 4 = 1.36758 × 106 J . For the force in this problem we have, W=
0 (a + b x − c x2 ) dx 1 1 = a x + b x2 − c x3 0 2 3 = (23100 N)(0.699 m) 1 + (12500 N/m)(0.699 m)2 2 1 − (24000 N/m2 )(0.699 m)3 3 = 16468.4 J . 005 10.0 points A force Fx acts on a particle. The force is related to the position of the particle by the formula Fx = (8.9 N/m3 ) x3 . Find the work done by this force on the particle as the particle moves from x = 1.9 m to x = 28 m. Correct answer: 1.36758 × 106 J. Explanation: Let : Fx = (8.9 N/m3 ) x3 , xi = 1.9 m , and xf = 28 m . The work done by a varying force is
x2 W=
x1 F · ds. All of the motion is in the ˆ direction so ds = ı dx, and
x2 W=
x1 F · dx ...
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This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.
 Spring '11
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