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A 5 kg rock is thrown down at 7 m/s from a height of 4 m above the floor...?
What will be its kinetic energy just before hitting the floor?
3 Answers
- 1 decade ago
Using the formula 1/2(mass)(velocity)^2 = the amount of kinetic energy would be....
1/2(5kg)(7m/s)^2
which would now be....
17.5^2
and the final answer would be
306.25 joules
Source(s): My science book. - ?Lv 51 decade ago
First you must find the final velocity due to gravity.
s = v0*t + (1/2)* a*t^2
4 = 7t + 4.9*t^2
0 = 4.9*t^2 + 7t -4
t = (-7 +/- sqrt (7^2 -4*4.9*(-4)))/(2*4.9)
t = (-7 +/- sqrt(49+78.4))/9.8
t = (-7 +/- sqrt(127.4))/9.8
t = (-7 +/-11.3)/9.8 but t can't be negative so
t = (-7 +11.3)/9.8 = 4.3/9.8 = 0.44 sec
v = v0 + at = 7 + 9.8*0.44 = 7 + 4.3 = 11.3 m/s
Now KE = (1/2)*m*v^2
= (1/2)*5*11.3^2
= 319J
