**Physics Problem: Acceleration Comparison***A force \( F_1 \) is applied to a mass \( M_1 \). If the same force \( F_1 \) is applied to a mass \( M_2 = 4M_1 \), the acceleration of mass \( M_1 \) is:*- ○ one-half the acceleration of the mass \( M_2 \).- ○ twice the acceleration of the mass \( M_2 \).- ○ the same as for the mass \( M_2 \).- ○ four times the acceleration of the mass \( M_2 \).- ○ one-fourth the acceleration of the mass \( M_2 \).**Explanation:**This problem requires understanding how acceleration is affected by mass when the same force is applied. According to Newton's second law, \( F = ma \), the acceleration (\( a \)) is inversely proportional to the mass (\( m \)) if the force (\( F \)) is constant. For mass \( M_2 = 4M_1 \), its acceleration will be \( \frac{1}{4} \) of that of \( M_1 \). Thus, the acceleration of mass \( M_1 \) is four times that of mass \( M_2 \).

Name: **Physics Problem: Acceleration Comparison***A force \( F_1 \) is app
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Description: **Physics Problem: Acceleration Comparison***A force \( F_1 \) is applied to a mass \( M_1 \). If the same force \( F_1 \) is applied to a mass \( M_2 = 4M_1 \), the acceleration of mass \( M_1 \) is:*- ○ one-half the acceleration of the mass \( M_2

Use Newton's second law for the mass M1 as, F1=M1a1 ...... 1 Here, a1 denotes the acceleration of the first mass.Use Newton's second law for the mass M2 as, F1=M2a2 Here, a2 denotes the acceleration of the second mass. F1=4M1a2 ...... 2 Now, equate the equations (1) and (2) as, M1a1=4M1a2a1=4a2 The acceleration of the first mass is four times the acceleration of the second mass. Thus, the fourth option is correct.

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A force Fj is applied to a mass M1. If the same force F1 is applied to a mass M2 = 4M¡ the acceleration of mass M, is: one-half the acceleration of the mass M. twice the acceleration of the mass Mɔ. the same as for the mass Mɔ. four times the acceleration of the mass M2. one-fourth the acceleration of the mass M2.

A force Fj is applied to a mass M1. If the same force F1 is applied to a mass M2 = 4M¡ the acceleration of mass M, is: one-half the acceleration of the mass M. twice the acceleration of the mass Mɔ. the same as for the mass Mɔ. four times the acceleration of the mass M2. one-fourth the acceleration of the mass M2.

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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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Newton's Second Law

Newton’s laws of motion describe the relationship between the motion of an object and forces acting on it. Newton’s law of motion has two types as follows:

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**Physics Problem: Acceleration Comparison**

*A force \( F_1 \) is applied to a mass \( M_1 \). If the same force \( F_1 \) is applied to a mass \( M_2 = 4M_1 \), the acceleration of mass \( M_1 \) is:*

- ○ one-half the acceleration of the mass \( M_2 \).
- ○ twice the acceleration of the mass \( M_2 \).
- ○ the same as for the mass \( M_2 \).
- ○ four times the acceleration of the mass \( M_2 \).
- ○ one-fourth the acceleration of the mass \( M_2 \).

**Explanation:**

This problem requires understanding how acceleration is affected by mass when the same force is applied. According to Newton's second law, \( F = ma \), the acceleration (\( a \)) is inversely proportional to the mass (\( m \)) if the force (\( F \)) is constant.

For mass \( M_2 = 4M_1 \), its acceleration will be \( \frac{1}{4} \) of that of \( M_1 \). Thus, the acceleration of mass \( M_1 \) is four times that of mass \( M_2 \).

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