A block of mass 3.5 kg, sliding on a horizontal plane, is released with a velocity of 3.3 m/s. The blocks 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 quadrupled

If the initial velocity were quadrupled the block would slide a distance of .

Explanation:

We are told the mass of the block is , the initial velocity in the first case is , the final velocity is and the distance at which the block stops in the first case is .

In the second case we are told the initial velocity is quadrupled.

Because the block doesn't move in the vertical direction the sum of forces is:

In the horizontal direction the only force we have is the force of kinetic friction is . We don't know the value of the coefficient of kinetic friction  but this force is the same in both cases.

We use that the change in kinetic energy is equal to the work done by :

Because  and is the same in both cases we have that:

is the distance the block will travel until stopping when the initial velocity is quadrupled. If we equal the above equation in the first and second case we have that:

If we manipulate the equation we get that:

f=ma newton's second law 7x8=56n

resonant frequency