Question

A block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed. a) if the mass of the block is doubled but the initial velocity is not changed what is the distance the block slides before stopping? b) if the initial velocity of the block is doubled to 2V0 but the mass is not changed, what is the distance the block slides before stopping?

Answer 1

Answer:

$a) \ d_2=d_1\\b) \ d_2=4d_1$

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

$E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}$

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.