Answer:
The block would have slid <u>12.6 m</u> if its initial velocity were increased by a factor of 2.8.
Explanation:
Given:
Mass of the block (m) = 3.6 kg
Initial velocity (u) = 1.7 m/s
Final velocity (v) = 0 m/s
Displacement (S) = 1.6 m
First we will find the acceleration of the block.
Using the equation of motion, we have:
![v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2aS%5C%5C%5C%5Ca%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2S%7D)
Now, plug in the given values and solve for 'a'. This gives,
![a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B0-1.7%5E2%7D%7B2%5Ctimes%201.6%7D%5C%5C%5C%5Ca%3D%5Cfrac%7B-2.89%7D%7B3.2%7D%3D0.903%5C%20m%2Fs%5E2)
The acceleration is negative as it is resisting the motion.
Now, the initial velocity is increased by a factor of 2.8. So,
New initial velocity = 2.8 × 1.7 = 4.76 m/s
Again using the same equation of motion and expressing the result in terms of 'S'. This gives,
![v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2aS%5C%5C%5C%5CS%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
Now, plug in the given values and solve for 'S'. This gives,
![S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B0-4.76%5E2%7D%7B2%5Ctimes%20-0.903%7D%5C%5C%5C%5CS%3D%5Cfrac%7B-22.6576%7D%7B-1.806%7D%3D12.6%5C%20m)
Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8