Question

A box of mass 41.9 kg is moving across a level floor at a speed of 3.53 m/sec. A kinetic friction force of 104.4 N is acting on the box directed against its motion. How far must the box move for the force of friction to bring it to rest?
Select one:
a. 4.07 m
b. 2.13 m
c. 17.3 m
d. 2.50 m
e. 3.31 m

Answer 1

Answer on Question #63309, Physics / Mechanics | Relativity

A box of mass $41.9\mathrm{kg}$ is moving across a level floor at a speed of $3.53\mathrm{m/sec}$ . A kinetic friction force of $104.4\mathrm{N}$ is acting on the box directed against its motion. How far must the box move for the force of friction to bring it to rest? Select one:

a. $4.07\mathrm{m}$

b. $2.13\mathrm{m}$

c. $17.3\mathrm{m}$

d. $2.50\mathrm{m}$

e. $3.31\mathrm{m}$

Find: s - ?

Given:

m=41.9 kg

$\mathrm{v_0 = 3.53~m / s}$

$F_{\text {frict}} = 104.4 \, \text{N}$

$\mathrm{g} = 9.8 \, \mathrm{m} / \mathrm{s}^{2}$

Solution:

Movement is delayed.

Distance:

$2\mathrm{as} = \mathrm{v}^{2} - \mathrm{v}_{0}^{2}(1),$

where $v = 0 \, \text{m/s}$

Of (1) $\Rightarrow$ $2\mathrm{as} = \mathrm{v}_0^2$ (2)

Of (2) $\Rightarrow$ s = $\frac{v_0^2}{2a}$ (3)

The body moves only under the influence of friction force (along OX):

$\mathrm{F_{frict} = ma(4)}$

Friction force:

$$\mathrm {F} _ {\text {f r i c t}} = \mu \mathrm {m g} (5),$$

where $\mu$ is coefficient of friction

Of (5) $\Rightarrow \mu = \frac{F_{\text{frict}}}{\text{mg}}$ (6)

Of (6) $\Rightarrow \mu = 0.2543$

(5) in (4): $\mu \mathrm{mg} = \mathrm{ma}(7)$

Of (7) $\Rightarrow a = \mu g$ (8)

Of (8) $\Rightarrow a = 2.4922 \, \text{m/s}^2$

Of (3) $\Rightarrow s = 2.5 \, \text{m}$

**Answer:**

d. $2.50\mathrm{m}$

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