Solution #4 summer 2007

# Solution #4 summer 2007 - Physics 2211 A Summer 2007 Quiz...

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Quiz #4 Solutions Summer 2007 G = 6 . 673 × 10 - 11 N · m 2 / kg 2 g Earth = 9 . 8m / s 2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I . (16 points) It is the end of the dog sled race, and a dog approaches the ﬁnish line over level ground pulling a 55kg sled. At 5.0m from the ﬁnish line, the sled is travelling at 6.0m/s. The coeﬃcient of kinetic friction between the sled and the ground is 0.18 in these last 5.0m. The tiring dog exerts a force, ~ F , on the sled whose magnitude decreases as the ﬁnish line is approached, according to ~ F ( x ) = 32 x where x is the distance from the sled to the ﬁnish line, and the force is in Newtons when the distance is in meters. What is the speed of the sled as it crosses the ﬁnish line? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Let the sled be the system, and use the Work-Energy Theorem: W ext + W nc = Δ K + Δ U With this choice of system, there are no potential en- ergy changes (Δ U = 0). The weight force and the normal force each do no work on the system, as they are always perpendicular to the displacement. The force exerted by the dog is an external force parallel to the displacement, and the force of friction is a non-conservative force anti-parallel to the displacement. Z d 0 F ( x )cos0 dx + Z d 0 f cos180 dx = 1 2 mv 2 f - 1 2 mv 2 i where d is the distance travelled, 5.0m. Note that the normal force is equal in magnitude to the weight in this situation, so the frictional force f = μ k n = μ k mg . So Z d 0 32 xdx - Z d 0 μ k mg dx = 32 x 3 / 2 3 / 2 d 0 - μ k mgx d 0 = 2 · 32 3 d 3 / 2 - μ k mgd = 1 2 mv 2 f - 1 2 mv 2 i Solving for the ﬁnal speed v f = r v 2 i + 4 · 32 3 m d 3 / 2 - 2 μ k gd At this point, it is worth noting that the “32” in the expression for the force exerted by the dog must have units of N / m 1 / 2 . v

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## Solution #4 summer 2007 - Physics 2211 A Summer 2007 Quiz...

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