139k views
16 votes
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

2 Answers

5 votes

Answer:

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 !

User Universal Grasp
by
7.9k points
3 votes

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.

User Jjujuma
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.