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A $$5.00kg$$ block is placed on top of a $$10.0kg$$ block (Fig. above). A horizontal force of $$45.0 N$$ is applied to the $$10kg$$ block, and the $$5.00kg$$ block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is $$0.200$$. (a) Draw a free-body diagram for each block and identify the actionreaction forces between the blocks. (b) Determine the tension in the string and the magnitude of the acceleration of the $$10.0kg$$ block.

Solution
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(a) The free-body diagrams are shown in the figure below.
$$f_{1}$$ and $$n_{1}$$ appear in both diagrams as action-reaction pairs.
(b) For the $$5.00kg$$ mass, Newton’s second law in the $$y$$ direction gives:
$$n_{1} = m_{1}g = (5.00 kg)(9.80 m/s^{2} ) = 49.0 N$$
In the $$x$$ direction,
$$f_{1} − T = 0$$
$$T = f_{1} = \mu mg = 0.200(5.00 kg)(9.80 m/s^{2} ) = 9.80 N$$
For the $$10.0kg$$ mass, Newton’s second law in the $$x$$ direction gives:
$$45.0 N − f_{1} − f_{2} = (10.0 kg)a$$
In the $$y$$ direction,
$$n_{2} − n_{1} − 98.0 N = 0$$
$$f_{2} = \mu n_{2} = \mu (n_{1} + 98.0 N) = 0.20(49.0 N + 98.0 N) = 29.4 N$$
$$45.0 N − 9.80 N − 29.4 N = (10.0 kg)a$$
$$a = 0.580 m/s^{2}$$

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