p08_002.pdf

2. (a) Noting that the vertical displacement is 10 . 0

−

1 . 5=8 . 5 m downward (same direction as

F g ),

Eq. 7-12 yields

W g = mgd cos φ =(2 . 00)(9 . 8)(8 . 5) cos 0 ◦ = 167 J .

(b) One approach (which is fairly trivial) is to use Eq. 8-1, but we feel it is instructive to instead

calculate this as ∆ U where U = mgy (with upwards understood to be the + y direction).

∆ U = mgy f −

mgy i =(2 . 00)(9 . 8)(1 . 5)

−

(2 . 00)(9 . 8)(10 . 0) =

−

167 J .

(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 = 100 J at the ground), so we

again obtain W g = 167 J.

(f) As a result of Eq. 8-1, we must again ﬁnd ∆ U =

−

W g =

−

167 J.

(g) With this new information (that U 0 = 100 J where y =0)wehave U i = mgy i + U 0 = 296 J.

(h) With this new information (that U 0 = 100 J where y =0)wehave U f = mgy f + U 0 = 129 J. We

can check part (f) by subtracting the new U i from this result.