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Kris is holding a 2-kg book 1.5m above the floor. What is the potential energy of the book?
Answers
Answer:
PE= mgh= 2×10× 1.5= 30J
Answer:
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.