you lift a 2.0 kg basketball from the floor, where its gravitational energy is zero. You raise it 2.4 meters, carry it horizontally to a window, and drop it out. It falls 12.0 meters. What is its gravitational potential energy when it hits the ground? If 20000 joules of work is done in pumping water up to a height of 12 meters, how much water is pumped?
Answers
Answer:
Hii
Explanation:
Noting that the vertical displacement is 10.0m–1.50m=8.50m
downward (same direction as
F
g
), Eq. 7−12 yields
W
g
=mgdcosϕ=(2.00kg)(9.80ms
2
)(8.50m)cos0
∘
=167J.
(b) One approach (which is fairly trivial) is to use Eq. 8−1, but we feel it is instructive to instead calculate this as ΔU where U=mgy (with upward understood to be the +y direction). The result is
ΔU=mg(y
f
−y
i
)=(2.00kg)(9.80ms
2
)(1.50m−10.0m)=−167J.
(c) In part (b) we used the fact that U
i
=mgy
i
=196J.
(d) In part (b), we also used the fact U
f
=mgy
f
=29J.
(e) The computation of W
g
does not use the new information (that U=100J at the ground), so we again obtain W
g
=167J.
(f) As a result of Eq. 8−1, we must again find ΔU=–W
g
=–167J.
(g) With this new information (that U
0
=100J where y=0) we have
U
i
=mgy
i
+U
0
=296J.
(h) With this new information (that U
0
=100J where y=0) we have
U
f
=mgy
f
+U
0
=129J.
We can check part (f) by subtracting the new U
i
from this result.