A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a distance of 1.6 m beyond the point where it was released. How far would the block have slid if its initial velocity were increased by a factor of 2.8

The block would have slid 12.6 m 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

im guessing the west of station b

there are no contact forces mentioned on the list of choices that you've provided.