Question

Repeat Problem 41 with the second 3.0-kg block replaced by a $5.0-\mathrm{kg}$ block moving to the right at $3.0 \mathrm{~m} / \mathrm{s}$.

Solution

VerifiedAnswered 1 year ago

Answered 1 year ago

Step 1

1 of 11**(a)** In this part of the problem, we are asked to calculate the total kinetic energy of the two blocks. The known values are given as follows:

$\begin{aligned} m_1 & = 3.0\text{ kg}\\ m_2 & = 5.0\text{ kg}\\ v_1 & = 5.0\mathrm{~\dfrac{m}{s}}\\ v_2 & = 3.0\mathrm{~\dfrac{m}{s}}\\ \end{aligned}$

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