Question

A block pushed along the floor with velocity v0xv0x slides a distance dd after the pushing force is removed. If the mass of the block is doubled but its initial velocity is not changed, what distance does the block slide before stopping?

Answer 1

Answer:

There is no change in distance

Explanation:

Given that block slides a distance dd with initial velocity $v_{ox}$$v_{ox}$

Work energy theorem states that Work done W is the difference between final kinetic energy $KE_{f}$ and initial kinetic energy $KE_{i}$

W = $KE_{f}$ -  $KE_{i}$  ...........(equation 1)

In the given problem, the work is done by frictional force

Hence the work done is given by

$W_{Friction}$ =  -$F_{Friction}$ x dd

$F_{Friction}$ is negative since it is resistive force and dd is the distance

$W_{Friction}$ = - μ mg X dd

where μ is coefficient of friction.

Also we know that Kinetic energy KE = $\frac{1}{2}$ $mv^{2}$

$KE_{f}$ = 0 since final velocity is zero

$KE_{i}$ = $\frac{1}{2}$ $m(v_{ox} v_{ox}) ^{2}$

Substituting the corresponding values equation 1 becomes

-μ mg x dd = 0 -  $\frac{1}{2}$ $m(v_{ox} v_{ox}) ^{2}$

dd= $\frac{(v_{ox}v_{ox} )^{2} }{2g}$x (1/μ) ........... (equation 2)

To find distance when mass of block is doubled and initial velocity is not changed:

Equation 2 shows that the distance dd is independent of mass, therefore there is no change in distance.