13. A force acts on a 2.7 kg object that moves along the x axis (see figure). If the velocity v̟ is 1.2 m/s at 1=0, what is the velocity at 1 = 4.0 s? K, (N) 16 a. 0.28 m/s b. 9.2 m/s c. 2.7 m/s d. 12 m/s e. 5.2 m/s
13. A force acts on a 2.7 kg object that moves along the x axis (see figure). If the velocity v̟ is 1.2 m/s at 1=0, what is the velocity at 1 = 4.0 s? K, (N) 16 a. 0.28 m/s b. 9.2 m/s c. 2.7 m/s d. 12 m/s e. 5.2 m/s
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:**Problem Statement:**
13. A force acts on a 2.7 kg object that moves along the x-axis (see figure). If the velocity \( v_x \) is 1.2 m/s at \( t = 0 \), what is the velocity at \( t = 4.0 \) s?
**Options:**
a. 0.28 m/s
b. 9.2 m/s
c. 2.7 m/s
d. 12 m/s
e. 5.2 m/s
**Graph Description:**
- The graph is a Force vs. Time (F vs. t) plot.
- The x-axis represents time \( t \) in seconds, ranging from 0 to 4 seconds.
- The y-axis represents force \( F_x \) in Newtons, ranging from -8 N to 16 N.
**Graph Details:**
- Force is -6 N at t = 0 s.
- Force increases linearly over time, intersecting the x-axis at approximately t = 1 s.
- At t = 4 s, the force \( F_x \) reaches 16 N.
**Analysis:**
To find the velocity at \( t = 4.0 \) s, apply the impulse-momentum theorem, which involves calculating the area under the force vs. time graph (impulse) and dividing by the object's mass to find the change in velocity. Add this change to the initial velocity to find the final velocity.
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