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Unformatted text preview: Solution for Practice Test 4 for Test 2 Solution to Practice Test Problem 4.1() Problem: Let’s think back to some of the data analysis we have done in lab. Let’s say we want to calibrate our force sensor using a spring scale. We want to pull on the force sensor with a force of 2 . 0N . If, when we hang a 250g object from the spring scale, it stretches by 5cm , how far do we need to extend the spring when we calibrate the sensor? Sensor Mass Solution First, we can use the data given to calculate the spring constant k . When the mass hangs from the spring, the two forces on the mass are balanced, or Σ vector F = mvectora = 0 So to balance the force of gravity acting downward on the . 250kg object, there must be a spring force upward of the same size: 0 = Σ vector F = vector F c s,o + vector F g e,o Taking up to be positive: 0 = + kx − mg k = mg x = (0 . 250kg)(9 . 8 m s 2 ) . 05m = 49 N m Now we can calculate the displacement needed to calibrate the force sensor: 2N = kx cal x cal = 2N k = 2N 49 N m ≈ . 04m Grading Key: 4 Points 1 point(s) : correct form for force for spring 1 point(s) : F spring balanced against mg 1 point(s) : results of balance subbed into force on spring to get stretch for found k 1 point(s) : correct answer with units (this includes converting everything correctly) Total Points for Problem: 4 Points Solution to Practice Test Problem 4.2() Problem: A hockey puck of inertia . 5kg travels at 10 m s to the right and collides with a lump of Silly Putty ( m = 0 . 1kg ) at rest. The combined lump of puck and putty moves at 3 . 5 m s to the right. 1 (a)What is the energy dissipated in the collision? (b)Is this collision isolated? Why or why not? (c)What would be the total convertible energy in this collision, assuming it was isolated? (Do you want to change your answer to (b)?) (d)If the system is not isolated, and the collision lasted . 1s , what is the average external force exerted on the system? (e)What is the impulse delivered to the puck in the collision? Solution to part (a) KE f + U f + E diss = KE i + U i + E source There is no source energy (i.e. nothing here can eat or burn fuel). We aren’t told that the surface has a slope to it, so gravitational potential energy does not change, and there are no springs or anything else like that involved. In a collision with no energy dissipation, KE f = KE i , or KE f − KE i = 0 . However, since this is an inelastic collision with obvious deformation of the Silly Putty (due to its squishyness), the change in kinetic energy will not be zero, and this value is the energy dissipated. KE f − KE i = Δ KE = − E diss Find the kinetic energies: KE i = 1 2 m 1 v 2 01 + 1 2 m 2 v 2 02 = 1 2 (0 . 5kg) parenleftBig 10 m s parenrightBig 2 + 0 = 25J KE f = 1 2 ( m 1 + m 2 ) v 2 f = 1 2 (0 . 6kg) parenleftBig 3 . 5 m s parenrightBig 2 = 3 . 7J Now compute Δ E diss : E diss = KE i − KE f = 25J − 3 . 7J = 21 . 3J Grading Key: Part (a) 4 Points 2 point(s) : E diss = − Δ KE . No forgiveness if use formula for isolated....
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This note was uploaded on 05/04/2008 for the course PHYS 2054 taught by Professor Stewart during the Spring '08 term at Arkansas.
 Spring '08
 Stewart
 Physics, Force

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