Practice 06 Solutions

# Practice 06 Solutions - practice 06 – GADHIA TEJAS –...

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Unformatted text preview: practice 06 – GADHIA, TEJAS – Due: Feb 21 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A child of mass 24 . 1 kg takes a ride on an irregularly curved slide of height 6 . 05 m, as in the figure below. The child starts from rest at the top. The acceleration of gravity is 9 . 8 m / s 2 . h w = m g n If a frictional force acts on the child, what is the magnitude of the mechanical energy dissipated by this force, if the speed of the child at the bottom is 8 . 4 m / s? Correct answer: 578 . 641 J (tolerance ± 1 %). Explanation: In this case Δ K ext is not equal to zero and mechanical energy is not constant. We can use the equation Δ K + Δ U = Δ K ext , to find the loss of kinetic energy due to fric- tion, assuming the final speed at the bottom is known Δ K ext = E f- E i = 1 2 mv 2 f- mg h = 1 2 (24 . 1 kg) (8 . 4 m / s) 2- (24 . 1 kg) (9 . 8 m / s 2 ) (6 . 05 m) =- 578 . 641 J | Δ K ext | = 578 . 641 J . Question 2 part 1 of 3 10 points Assume: The pulleys are weightless, the rope does not stretch, and the system moves at a constant speed which is slow enough that the kinetic energy is negligible. A weight of 526 N is raised by a two-pulley arrangement as shown in the figure. m d F Δ x Note: Figure may not be drawn to scale. How much work is done by the agent (force vector F ) to raise the weight by a vertical distance of 12 m? Correct answer: 6312 J (tolerance ± 1 %). Explanation: Basic Concepts: W = vector F · vectors U grav . = mg h vector F net = summationdisplay i vector F i Solution: The work done by the force F results in the change of potential energy of the system. The final energy is the potential energy PE final = mg Δ x ; the initial energy is zero. Thus Δ E = mg Δ x = (526 N) (12 m) = 6312 J . Question 3 part 2 of 3 10 points Which of the following is the SI unit of the force [ F ]? 1. [ F ] = m/s 2 practice 06 – GADHIA, TEJAS – Due: Feb 21 2007, 4:00 am 2 2. [ F ] = W 3. [ F ] = N m 4. [ F ] = N/s 5. [ F ] = kg · m/s 6. [ F ] = J 7. [ F ] = m/kg · s 8. [ F ] = kg · m/s 2 correct 9. [ F ] = J s 10. [ F ] = kg Explanation: The SI unit of force [ F ] is kg · m/s 2 , called a Newton (N). Question 4 part 3 of 3 10 points The change in potential energy U of the weight is 1. U = mg d 2. U = 1 2 (2 F- mg ) d 3. U = 2 ( F + mg ) d 4. U = mg d 2 correct 5. U = ( F- mg ) d 6. U = 1 2 ( F- mg ) d 7. U = ( F- 2 mg ) d 8. U = 3 2 F d 9. U = 2 mg d 10. U = 2 ( F- mg ) d Explanation: If we consider the forces on the pulley we have T mg T As the weight is lifted with no acceleration we have, by force balance, 2 T = mg , or T = F = mg 2 . If d is the distance through which the force F acts, then the work done is W = F d = mg 2 d. Using the result of Part 1, W = mg Δ x = mg 2 d, so d = 2 Δ x d 2 = Δ x....
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Practice 06 Solutions - practice 06 – GADHIA TEJAS –...

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