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BARSIC [14]
1 year ago
15

Multiple-Object Accelerating Systems: As shown in the figure, two blocks are connected by a very light string, and the upper blo

ck is pulled upward by a different string. The masses of the upper and lower blocks are m1 = 300 g and m2 = 240 g, respectively. The tension in the upper string is equal to 6.6 N.a) Find the acceleration of the system.b) Find the tension in the lower rope.

Physics
1 answer:
djverab [1.8K]1 year ago
3 0

Part A.

To get the acceleration of the system we consider the two blocks as a single mass. For this situation we have, from Newton's second law, that:

T-W=(m_1+m_2)a

where T is the tension in the upper sting and W is the weight of the system. Solving the equation for a we have:

\begin{gathered} 6.6-(0.3+0.24)(9.8)=(0.3+0.24)a \\ a=\frac{6.6-(0.3+0.24)(9.8)}{(0.3+0.24)} \\ a=2.42 \end{gathered}

Therefore the acceleration of the system is 2.42 meters per second per second.

Part B.

Now, that we have the acceleration of the system we analyze the lower block individually; for this block the equation of motion is:

T^{\prime}-W^{\prime}=m_2a

where T' is the tension in the lower rope, W' is the weight of the lower block and m2 is its mass. Solving for the tension we have that:

\begin{gathered} T^{\prime}=(0.24)(9.8)+(0.24)(2.42) \\ T^{\prime}=2.93 \end{gathered}

Therefore the tension in the lower rope is 2.93 N

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Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

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Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

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<em>-Hope This Helps!</em>

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