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Answer:
A 5.00- kg block is placed on top of a 10.0 -kg block (Fig. P5.68). A horizontal force of 45.0 N is applied to the 10.0-kg block, and the 5.00- kg block is tied to the wall. The coefficient of kinetic friction between all surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action-reaction forces between the blocks.
(b) Determine the tension in the string and the magnitude of the acceleration of the 10.0-kg block.
The horizontal acceleration of the block is 5 m/s².
To calculate the horizontal acceleration on the block, we use the formula below.
Formula:
- ma = (F-F')............... Equation 1
Where:
- m = mass of the block
- a = Horizontal acceleration of the block
- F = Horizontal force exerted on the string
- F' = Frictional force
Make "a" the subject of the equation.
- a = (F-F')/m............... Equation 2
Substitute these values into equation 2
- F = 30 N
- F' = 5 N
- m = 5.0 kg
Substitute these values into equation 2
- a = (30-5)/5
- a = 25/5
- a = 5 m/s²
Hence the horizontal acceleration of the block is 5 m/s².
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