HW4-phy2048-Spring-2014-solution

7p16 the change in kinetic energy can be written as

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Unformatted text preview: 14 HW4 ! (f) The force F is again opposite to the direction of motion (so the angle φ is again 180°) so that Eq. 7- 7 leads to W = − F Δx = −5.8 × 104 J. The fact that this agrees with the result of part (c) provides insight into the concept of work. 7P16. The change in kinetic energy can be written as 1 1 Δ K = m(v 2 − vi2 ) = m(2aΔ x) = maΔ x f 2 2 where we have used v 2 = vi2 + 2aΔ x from Table 2- 1. From the figure, we see that f Δ K = (0 − 30) J = − 30 J when Δ x = + 5 m . The acceleration can then be obtained as ΔK (− 30 J) a= = = − 0.75 m/s 2 . mΔ x (8.0 kg)(5.0 m) The negative sign indicates that the mass is decelerating. From the figure, we also see that when x = 5 m the kinetic energy becomes zero, implying that the mass comes to rest momentarily. Thus, 2 v0 = v 2 − 2aΔ x = 0 − 2(− 0.75 m/s 2 )(5.0 m) = 7.5 m 2 /s 2 , or v0 = 2.7 m/s . The speed of the object when x = −3.0 m is 2 v = v0 + 2aΔ x = 7.5 m2 /s 2 + 2(− 0.75 m/s 2 )( −3.0 m) = 12 m/s = 3.5 m/s . 7P22....
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This document was uploaded on 03/24/2014.

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