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Chapter 4: Problem 32

4.32 A hanging mass, \(M_{1}=0.50 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. Find the magnitude of the acceleration, \(a,\) of \(M_{2}\)

Short Answer

Expert verified
Answer: The magnitude of the acceleration of mass \(M_2\) is approximately \(4.9\, \text{m/s}^2\).

Step by step solution

01

Identify the forces acting on each mass

First, we need to identify the forces acting on each mass. For mass \(M_1\), there is the gravitational force acting downward, which equals \(M_1g\). For mass \(M_2\), there is the force due to the tension in the string, \(T\).
02

Apply Newton's second law to each mass

Newton's second law states that the net force acting on an object equals its mass times acceleration (\(F_{net} = ma\)). For the vertical direction, the net force acting on mass \(M_1\) is given by the tension in the string minus the gravitational force (\(M_1g\)): \(F_{net1} = T - M_1g\). Similarly, for the horizontal direction, the net force acting on mass \(M_2\) is given by the tension in the string: \(F_{net2} = T\). Note that the two masses have the same acceleration because they are connected by the string. We can write two equations from Newton's second law for the two masses: For mass \(M_1\): \(T - M_1g = M_1a\) For mass \(M_2\): \(T = M_2a\)
03

Eliminate the unknown tension T

We have two equations and two unknowns, tension \(T\) and acceleration \(a\). We can eliminate the unknown tension by solving the equation for mass \(M_2\) for \(T\) and substituting into the equation for mass \(M_1\): \(T = M_2a\) Substitute into the equation for mass \(M_1\): \(M_2a - M_1g = M_1a\)
04

Solve for acceleration a

Now we have a single equation and one unknown, acceleration \(a\). Rearrange the equation to solve for \(a\): \(M_2a - M_1a = M_1g\) \((M_2 - M_1)a = M_1g\) \(a = \dfrac{M_1g}{M_2 - M_1}\) Now, we can substitute in the given values for the masses and the gravitational acceleration \(g = 9.81\,\text{m/s}^2\): \(a = \dfrac{(0.50 \,\text{kg})(9.81 \,\text{m/s}^2)}{1.50 \,\text{kg} - 0.50 \,\text{kg}}\) \(a = \dfrac{(0.50 \,\text{kg})(9.81 \,\text{m/s}^2)}{1.00 \,\text{kg}}\) \(a = 4.905 \,\text{m/s}^2\) So, the magnitude of the acceleration of mass \(M_2\) is approximately \(4.9\, \text{m/s}^2\).

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Most popular questions from this chapter

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