Calculate the speed of the object when it is at x=2m (b) Does the object reach x=3m ? ____ Yes ____ No (c) Is the force acting on the object positive, negative, or zero between x=1m and x=2m ? _____ Positive _____ Negative _____ Zero
Calculate the speed of the object when it is at x=2m (b) Does the object reach x=3m ? ____ Yes ____ No (c) Is the force acting on the object positive, negative, or zero between x=1m and x=2m ? _____ Positive _____ Negative _____ Zero
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(a) Calculate the speed of the object when it is at x=2m
(b) Does the object reach x=3m ?
____ Yes ____ No
(c) Is the force acting on the object positive, negative, or zero between x=1m and x=2m ?
_____ Positive _____ Negative _____ Zero
(d) Calculate the acceleration of the object when it is at x=1.5m.
(e) On the graph above, draw a dot at the position where the maximum negative force is exerted on the
object.
Provide your reasoning.
(f) When the object passes the origin, it is moving in the -direction with a kinetic energy of 20 .
Will the object ever return to the origin?
____ Yes ____ No
Provide your reasoning.
![**Graph and Function Explanation**
**Graph Description:**
- The graph displays the potential energy \( U(x) \) in joules (J) on the vertical axis against the position \( x \) in meters (m) on the horizontal axis.
- The plotted curve shows the potential energy changing as the position of the object varies from 0 to 4 meters.
- The potential energy starts at 0 J when \( x = 0 \), rises to about 33 J at around \( x = 2.4 \) m, and then decreases to 4 J at \( x = 4 \) m.
**Text Explanation:**
An object of mass 4.0 kg is moving along the x-axis. A conservative force is exerted on the object. The potential energy \( U \) associated with this force is given as a function of position \( x \) by the equation:
\[
U(x) = Kx^3 + Lx^2 + Mx + N
\]
where \( K = -\frac{3}{4} \), \( L = 10 \), \( M = -\frac{5}{2} \), and \( N = 4 \) J. This function is shown in the graph provided. The kinetic energy of the object when \( x = 0 \) is 30 J.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3083a394-edcd-43cd-8705-3794807b52ec%2F9e8c5715-205c-432a-a8f6-8d1e57832020%2Fi2zq1c_processed.png&w=3840&q=75)
Transcribed Image Text:**Graph and Function Explanation**
**Graph Description:**
- The graph displays the potential energy \( U(x) \) in joules (J) on the vertical axis against the position \( x \) in meters (m) on the horizontal axis.
- The plotted curve shows the potential energy changing as the position of the object varies from 0 to 4 meters.
- The potential energy starts at 0 J when \( x = 0 \), rises to about 33 J at around \( x = 2.4 \) m, and then decreases to 4 J at \( x = 4 \) m.
**Text Explanation:**
An object of mass 4.0 kg is moving along the x-axis. A conservative force is exerted on the object. The potential energy \( U \) associated with this force is given as a function of position \( x \) by the equation:
\[
U(x) = Kx^3 + Lx^2 + Mx + N
\]
where \( K = -\frac{3}{4} \), \( L = 10 \), \( M = -\frac{5}{2} \), and \( N = 4 \) J. This function is shown in the graph provided. The kinetic energy of the object when \( x = 0 \) is 30 J.
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