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Unformatted text preview: reyes (mr29667) – Homework 6 – chelikowsky – (59005) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Batman, whose mass is 78 . 8 kg, is holding on to the free end of a 15 . 8 m rope, the other end of which is fixed to a tree limb above. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough that he can reach a ledge when the rope makes a 77 . 7 ◦ angle with the vertical. The acceleration of gravity is 9 . 8 m / s 2 . How much work was done against the force of gravity in this maneuver? Correct answer: 9602 . 12 J. Explanation: The work done is given by the expression, W = m g Δ y . The change in height, Δ y , is Δ y = ℓ ℓ cos θ . Thus W = mg ( ℓ ℓ cos θ ) = mg ℓ (1 cos θ ) = (78 . 8 kg) (9 . 8 m / s 2 ) × (15 . 8 m) (1 cos 77 . 7 ◦ ) = 9602 . 12 J . 002 10.0 points A force F is exerted by a broom handle on the head of the broom, which has a mass m . The handle is at an angle θ to the horizontal, as shown below. F θ What is the work done by the force on the head of the broom as it moves a distance d across a horizontal floor? 1. W = F d cos θ correct 2. W = F d sin θ 3. W = F md sin θ 4. W = F m cos θ 5. W = F m tan θ Explanation: By definition, the work done by a force is given by the formula W = vector F · vector d = F d cos θ, where vector d is the displacement vector, and θ is the angle between vector F and vector d . Simply apply the definition, the work is W = F d cos θ . 003 10.0 points You drag a(n) 19 . 3 kg steamer trunk over a rough surface by a constant force of 162 N acting at an angle of 30 . 1 ◦ above the horizon tal. You move the trunk over a distance of 7 . 99 m in a straight line, and the coefficient of kinetic friction is 0 . 186. The acceleration of gravity is 9 . 8 m / s 2 . 19 . 3 kg μ = 0 . 186 1 6 2 N 3 . 1 ◦ How much is the work done on the block by the net force? Correct answer: 959 . 487 J. Explanation: reyes (mr29667) – Homework 6 – chelikowsky – (59005) 2 F θ mg n f k Work is W = vector F · vectors , where vectors is the distance traveled. In this problem vectors = 5ˆ x is only in the ˆ x direction. W F = vector F x · vectors x = F s x cos θ = (162 N) (7 . 99 m) cos30 . 1 ◦ = 1119 . 83 J . To find the frictional force, F friction = μ N , we need to find N from vertical force balance. Note: N is in the same direction as the y component of F and opposite the force of gravity. Thus F sin θ + N = mg , so that N = mg F sin θ . Thus the friction force is vector F friction = μ N ˆ x = μ ( mg F sin θ ) ˆ x , and the work done by friction is W μ = vector F friction · vectors = F f   s  = μ ( mg F sin θ ) s x = . 186 bracketleftBig (19 . 3 kg) (9 . 8 m / s 2 ) (162 N) sin30 . 1 ◦ bracketrightBig (7 . 99 m) = 160 . 347 J ....
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 Fall '09
 Chelikowsky
 Energy, Force, Friction, Potential Energy, Correct Answer, Reyes

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