Hwk16 - Solution for Long-answer Homework 16 Work problems Solution to Long-answer Homework Problem 16.1(Lizard Acceleration Problem A lizard jumps

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Unformatted text preview: Solution for Long-answer Homework 16 Work problems Solution to Long-answer Homework Problem 16.1(Lizard Acceleration...) Problem: A lizard jumps, pushing with its legs against the ground. (a)Assume the lizard’s inertia is . 05kg . Its center of mass starts out . 5cm above the floor, and raises 1 . 0cm to 1 . 5cm while it is pushing against the floor. At that height, its feet lose contact with the floor and it continues to travel up to a height of . 15m . What is the magnitude of the average force exerted by the floor on the lizard? (b)What was the lizard’s acceleration during the launch? Solution to Part (a) One way to do this is to use kinematics to find the lizard’s acceleration, but I’m going to do it the easy way. However, we have to think a little, because there is an extra step in the calculation to figure out how fast the lizard was going when it left the ground. When the lizard leaves the ground, it has enough kinetic energy to take it up to 15cm . At 15cm , the lizard will have zero velocity, and it started at 1 . 5cm with an initial velocity v i = v launch . Since no outside forces are acting on the lizard during its flight (if we put the Earth in the system and conserve energy), Δ U =- Δ K mgh f- mgh i =- (0- 1 2 mv 2 launch ) = 1 2 mv 2 launch Now, let’s see how much force we need to get that launch velocity. For the first part of the motion, while it is pushing, there is source energy being converted to kinetic energy. There is no work done, because the floor doesn’t move, so the point of contact doesn’t move. Remember, if the point of contact doesn’t move, any forces just let you convert energy from one kind to another inside the system. We only want to worry about the center of mass of the lizard, so lets take the Earth out of the system for this part of the calculation. Our starting kinetic energy is zero, and our final is for this part of the motion is what we start out with as we leave the launch. F net Δ x cm = Δ K F net Δ x cm = 1 2 mv 2 f- 1 2 mv 2 i where v launch = v f in this situation, and v i = 0 (before the lizard jumps), F net = 1 2 mv 2 launch Δ x cm = ( mgh f- mgh i ) Δ x cm and so F net = (0 . 05kg)(9 . 81 m s 2 )(0 . 135m) (0 . 01m) = 6 . 6N Now, since the net force is upward, in the direction of the force due to the floor, and gravity is opposite that: vector F net = vector F c floor,l + vector F g e,l = | vector F c floor,l | - | vector F g e,l | Rearranging and substituting in to find the average force exerted by the floor, | vector F c floor,l | = | vector F net | + | vector F g e,l | = 6 . 6N + mg = 6 . 6N + (0 . 05kg)(9 . 81 m s 2 ) = 7 . 1N Grading Key: Part (a) 4 Points 1 Solution to Part (b) F net = ma a = F net m = 132 . 4 m s 2 Grading Key: Part (b) 2 Points Total Points for Problem: 6 Points...
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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.

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Hwk16 - Solution for Long-answer Homework 16 Work problems Solution to Long-answer Homework Problem 16.1(Lizard Acceleration Problem A lizard jumps

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