Answer:
<h3>We need to apply a force of 15 Netwons in order to move the block at a constant velocity.</h3><h3>For the second context the acceleration is 4.90 meters per square second.</h3>
Explanation:
Notice that the problem is about a block, which we need to move with constant velocity, that means there's no acceleration, because the velocity doesn't variate.
We use Newton's Second Law ![\Sigma F = ma](https://tex.z-dn.net/?f=%5CSigma%20F%20%3D%20ma)
Where
and
(friction coefficient).
In this case, we know that
, because there's no acceleration.
![F - f_{\mu}=0](https://tex.z-dn.net/?f=F%20-%20f_%7B%5Cmu%7D%3D0)
Where, ![f_{\mu} = \mu N = \mu W = 0.5(30)=15 \ N](https://tex.z-dn.net/?f=f_%7B%5Cmu%7D%20%3D%20%5Cmu%20N%20%3D%20%5Cmu%20W%20%3D%200.5%2830%29%3D15%20%5C%20N)
Therefore,
, which means we need to apply a force of 15 Netwons in order to move the block at a constant velocity.
Now, in case we use double force
, we need to find the mass of the block using ![g= 9.81 \ m/s^{2}](https://tex.z-dn.net/?f=g%3D%209.81%20%5C%20m%2Fs%5E%7B2%7D)
![mg=30\\m=\frac{30}{9.81} \approx 3.06 \ kg](https://tex.z-dn.net/?f=mg%3D30%5C%5Cm%3D%5Cfrac%7B30%7D%7B9.81%7D%20%5Capprox%203.06%20%5C%20kg)
Now, according to Newton's Second Law, we have
![F - f_{\mu}=ma\\](https://tex.z-dn.net/?f=F%20-%20f_%7B%5Cmu%7D%3Dma%5C%5C)
Replacing all values and solving for
, we have
![30-15=3.06a\\a=\frac{15}{3.06} \approx 4.90 \ m/s^{2}](https://tex.z-dn.net/?f=30-15%3D3.06a%5C%5Ca%3D%5Cfrac%7B15%7D%7B3.06%7D%20%5Capprox%204.90%20%5C%20m%2Fs%5E%7B2%7D)
Therefore, for the second context the acceleration is 4.90 meters per square second.