Question

A 5.0-kg object is moving with speed 2.0 m/s. A 1.25-kg object is moving with speed 4.0 m/s Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping

Answer 1

Answer:

They stop in the same distance

Explanation:

f friction is the same for both and must overcome the KE of each object to bring each to rest.

f friction * d   = 1/2 mv^2    for each object :

f * d1 = 1/2 m1 v1^2                      f * d2 = 1/2 m2 v2^2

       =  1/2 5 (2 )^2                               = 1/2 1.25 (4)^2

Object 1  KE = 10  j                       Object 2 KE =    10 j

 

f d1 / f d2   =    10 / 10      ( f is the same for both)

  d1 / d2 = 1                    so  d1 = d2    they stop in the same distance !

 

Answer 2

Acceleration is constant, so we can use the following kinematic equation and Newton's second law.

${v_f}^2 - {v_i}^2 = 2a\Delta x \implies -{v_i}^2 = -2\dfrac{F}{m}\Delta x \implies {v_i}^2 = 2\dfrac{F}{m} \Delta x$

where $v_i$ and $v_f$ are initial/final velocities, $a$ is acceleration, $\Delta x$ is displacment, $F$ is force, and $m$ is mass. Since the objects are coming to rest, the acceleration opposes the direction of motion and is negative.

Solve for $\Delta x$.

$\Delta x = -\dfrac{m {v_i}^2}{2F}$

Compute the displacements of both objects.

$\Delta x_{5.0\text{-kg}} = -\dfrac{(5.0\,\mathrm{kg}) \left(2.0\frac{\rm m}{\rm s}\right)^2}{2F} = -\dfrac{10}F \dfrac{\rm m}{\rm N}$

$\Delta x_{1.25\text{-kg}} = -\dfrac{(1.25\,\mathrm{kg}) \left(4.0\frac{\rm m}{\rm s}\right)^2}{2F} = -\dfrac{10}F \dfrac{\rm m}{\rm N}$

Then both objects are displaced by the same amount.