At the moment when a shot putter releases a 5.00 kg shot, the shot is 3.00 m abo
ID: 1308184 • Letter: A
Question
At the moment when a shot putter releases a 5.00 kg shot, the shot is 3.00 m above the ground and travelling at 15.0 m/s. It reaches a maximum height of 14.5 m above the ground and then falls to the ground. If air resistance is negligible,
a) What was the potential energy of the shot as it left the hand relative to the ground?
b) What was the kinetic energy of the shot as it left the hand?
c) What was the total energy of the shot as it left the hand?
d) What was the total energy of the shot as it reached its maximum height?
e) What was the potential energy of the shot at its maximum height?
f) What was the kinetic energy of the shot at its maximum height?
g) What was the kinetic energy of the shot just as it struck the ground? [
a. 147 J b. 563 J c. 710. J d. 710. J e. 710. J f. 0 g. 710. J]
Explanation / Answer
a). PE= mgh
=5*3*9.8
= 147 J
b) KE= 1/2*m*V^2
=1/2*5*15^2
=562.5
= 563 J
c).TE= PE+KE
= 563+147
=710 J
d). when it reached its maximum height KE=0
and PE= mgh
= 5*9.8*14.5
=710 J
e). PE= mgh
= 710 J
f). KE=0 as velocity is zero
g). since total energy is conserved and on reaching ground its PE=0
so KE= TE
= 710 J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.