y mo-uploaded %2fohiou%217L31 941 dbc6 junshuo Zhang, (Student -section: 102) Ma
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y mo-uploaded %2fohiou%217L31 941 dbc6 junshuo Zhang, (Student -section: 102) Main Menu Contents Grades PHYS 2051/Spring 2016 Course Contents » Pre-class assignment for Friday Wk 5 Kinetic energy Kinetic energy Due In 11 hours, 32 minutes Due in 11 hours, 32 minutes Find the kinetic energy in joules of a 0.120 kg baseball moving with a speed of 35.0 m/s. 7.350x101 You are cor Your receipt no. is 154-31() Find the kinetic energy in joules of a 65.0 kg jogger running at a steady pace of 10.00 min/mi. 1.17 t Answer Incorrect. Tries 1/10 Previous Tries Threaded View NEW Units Jacob Whitehead Reply (Sat Feb 6 10:44:15 pm 2016 (EST)) Just in case this throws anybody else off, the units min/mi means that the runner runs exactly 1 mile in x minutes. NEW Re: Units Anonymous 2 Reply (Thu Feb 11 02:32:03 pm 2016 (EST) Thank you! My settings for this discussion: 1. Display - All posts 2. Not new- Once marked not NEW Other Views... Change splay - All posts 2. Not new - Once marked not NEWExplanation / Answer
Here,
for the kinetic energy of baseball
Kinetic energy = 0.5 m * v^2
Kinetic energy = 0.5 * 0.120 * 35^2
Kinetic energy = 73.5 J
the kinetic energy of baseball is 73.5 J
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for the jogger
speed = 10 min/ml
speed = 1/10 mil/min = 60/10 mil/hr
speed = 6 * 1609 m/3600 s
speed = 2.68 m/s
Kinetic energy = 0.5 * 65 * 2.68^2
Kinetic energy = 233.71 J
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let the final velocity is vf
inintial velocity , vi = 1.72 *10^7 m/s
distance ,d = 3.88 cm = 0.0388 m
Using third equation of motion
vf^2 - vi^2 = 2 *a * d
vf^2 - (1.72 *10^7)^2 = 2 * 3.63 *10^15 * 0.0388
solving for vf
vf = 2.403 *10^3 m/s
the final speed is 2.403 *10^3 m/s
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increase in kinetic energy = work done by force
increase in kinetic energy = m * a * d
increase in kinetic energy = 1.67 *10^-27 * 3.63 *10^15 * 0.0388
increase in kinetic energy = 2.35 *10^-13 J
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