Problem 1 For this problem, your \"system\" will be a dog jumping up to catch a
ID: 1780060 • Letter: P
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Problem 1 For this problem, your "system" will be a dog jumping up to catch a frisbee that is flying overhead. Take the dog's mass to be 60 kg, and assume that when he crouches down for his jump his center of mass is only 0.3m above the ground. When he jumps, at the moment his feet leave the ground (so he can no longer push), his center of mass is 0.8m off the ground. Assume that he can push against the ground with a force 4 times the magnitude of his own weight. dog (a)Find the value of the net external force on the dog as he is pushing off the ground. Make sure to draw a free body diagram on the figure to the right for the dog during this time (while his feet are in contact with the ground). b)What objects would you need to draw to include the force pairs of the forces drawn on the dog? (c)Assuming the dog was at rest when it started pushing, what is its kinetic energy when its feet leave the ground? (d)How much work did the ground do on the dog (write this down!)? Carefully draw a work-energy diagram on the figure to the right showing how energy was transferred during the push for the defined system (the dog). Assume the dog's muscles are not perfectly efficient. (e)Assume the dog catches the frisbee if his center of mass | AK AU Ediss 4urce Work reaches 3m above the ground. Does he make it?Explanation / Answer
a) When the dog applies a force of 4g on the ground the normal reaction from the ground is 4mg
Forces acting on the dog when at rest are
1. its own weight mg - down
2. Normal reaction from the ground - 4mg
Netx extrenal force on the dog = 3mg = 3*60*9.8 = 1764 N , upward
b. dog and the ground are objects to explain the force pairs
c. When dog is at rest CM is at 0.3m ; u=0
when just left the ground CM is at 0.8m ; v = ?
v2 = 2gh = 2*9.8*0.5 = 9.8
KE = 0.5mv2 = 30*9.8 = 294J
v = 3.13 m/s
d. Force exterted by the ground = 3mg
displacment = 0.5 m
work done = 3*60*9.8*0.5 = 882 J
e. CM at h = 0.8m - intial height
initial vel u = 3.13 m
max, height raise h = sqrt(v/2g) = sqrt(3.132/2*9.8).
= 0.71 m
max. height of CM = 0.8 + 0.71 = 1.51m
it cannot catch the frisbee
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