Question

question 4 (1 point)
a total force t acts on a mass m and produces an acceleration a. what acceleration results if a total force 2t acts on mass 4m?
4a
8a
a/2
2a

Explanation:

Step1: Recall Newton's Second Law

Newton's second law states that \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration. For the first case, we have \( T = M \cdot a \).

Step2: Apply Newton's Second Law to the second case

In the second case, the force is \( 2T \) and the mass is \( 4M \). Let the new acceleration be \( a' \). So, \( 2T = 4M \cdot a' \).

Step3: Substitute \( T = M \cdot a \) into the second equation

From the first equation, \( T = Ma \), so substitute \( T \) in the second equation: \( 2(Ma) = 4M \cdot a' \).

Step4: Solve for \( a' \)

Simplify the equation: \( 2Ma = 4M \cdot a' \). Divide both sides by \( 2M \): \( a = 2a' \)? Wait, no, wait. Wait, let's do it again. \( 2T = 4M a' \), and \( T = Ma \), so substitute \( T \): \( 2(Ma) = 4M a' \). Then, \( 2Ma = 4M a' \). Divide both sides by \( 2M \): \( a = 2a' \)? No, that's not right. Wait, divide both sides by \( 4M \): \( a' = \frac{2T}{4M} \). But \( T = Ma \), so substitute \( T \): \( a' = \frac{2(Ma)}{4M} \). The \( M \) cancels out: \( a' = \frac{2a}{4} = \frac{a}{2} \). Wait, no, wait, \( 2T = 4M a' \), \( T = Ma \), so \( 2(Ma) = 4M a' \). Then, \( 2Ma = 4M a' \). Divide both sides by \( 2M \): \( a = 2a' \)? No, that's incorrect. Wait, let's do the algebra again. Starting over:

From \( T = M a \), so \( a = \frac{T}{M} \).

For the second case, \( F = 2T \), \( m = 4M \), so \( a' = \frac{F}{m} = \frac{2T}{4M} = \frac{T}{2M} \).

But since \( a = \frac{T}{M} \), then \( \frac{T}{2M} = \frac{a}{2} \). So the new acceleration \( a' = \frac{a}{2} \).

Answer:

\( \frac{a}{2} \) (which corresponds to the option "a/2")