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

They stop in the same distance

Step-by-step 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(F)/(m)\Delta x \implies {v_i}^2 = 2(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 = -\frac{m {v_i}^2}{2F}`

Compute the displacements of both objects.

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

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

Then both objects are displaced by the same amount.