The following graph shows a plot of food energy energy the the hardiche function of position, the Chegy med betensial
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Figure 8−50 shows a plot of potential energy U versus position x of a 0.90kg particle that can travel only along an x axis. (Non conservative forces are not involved.) Three values are U
A
=15.0J,U
B
=35.0J, and U
C
=45.0J. The particle is released at x=4.5m with an initial speed of 7.0m/s, headed in the negative x direction.(a) If the particle can reach x=1.0m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x=4.0m? Suppose, instead, the particle is headed in the positive x direction when it is released at x=4.5m at speed 7.0m/s. (d) If the particle can reach x=7.0m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and(f) direction of the force on the particle as it begins to move to the right of x=5.0m?
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Solution
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From the figure, we see that at x=4.5m, the potential energy is U
1
=15J. If the speed is v=7.0m/s, then the kinetic energy is
K
1
=mv
2
/2=(0.90kg)(7.0m/s)
2
/2=22J.
The total energy is E
1
=U
1
+K
1
=(15+22)J=37J.
(a) At x=1.0m, the potential energy is U
2
=35J. By energy conservation, we have K
2
=2.0J>0. This means that the particle can reach there with a corresponding speed
v
2
=
m
2K
2
=
0.90kg
2(2.0J)
=2.1m/s.
(b) The force acting on the particle is related to the potential energy by the negative of the slope:
F
x
=−
Δx
ΔU
From the figure we have F
x
=−
2m−4m
35J−15J
=+10N.
(c) Since the magnitude F
x
>0 , the force points in the +x direction.
(d) At x=7.0m, the potential energy is U
3
=45J, 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 6m, the potential energy is given by
U(x)=15+30(x−5), 5≤x≤6.
Thus, the turning point is found by solving 37=15+30(x−5), which yields x=5.7m.
(e) At x=5.0m, the force acting on the particle is
F
x
=−
Δx
ΔU
=−
(6−5)m
(45−15)J
=−30N.
The magnitude is ∣F
x
∣=30N.
(f) The fact that F
x
<0 indicated that the force points in the –x direction.
Explanation:
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