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:

Now, plug in the given values and solve for 'a'. This gives,

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,

Now, plug in the given values and solve for 'S'. This gives,

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8