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(b) The force acting on the particle is related to the potential energy by the negative of the
slope:
x
U
F
x
Δ
=−
Δ
From the figure we have
35 J 15 J
10 N
2 m 4 m
x
F
−
=+
−
.
(c) Since the magnitude
0
x
F
>
, the force points in the +
x
direction.
(d) At
x
= 7.0m, the potential energy is
U
3
= 45 J which exceeds the initial total energy
E
1
.
Thus, the particle can never reach there. At the turning point, the kinetic energy is zero.
Between
x
= 5 and 6 m, the potential energy is given by
( ) 15 30(
5),
5
6.
Ux
x
x
−
≤
≤
Thus, the turning point is found by solving 37 15 30(
5)
x
−
, which yields
x
= 5.7 m.
(e) At x =5.0 m, the force acting on the particle is
(45 15) J
30 N
(6 5) m
x
U
F
x
Δ−
The magnitude is |

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