Question

exit ticket
a 25 kg wood box is sliding across a wood surface, experiencing a coefficient
of kinetic friction of 0.20. the box is experiencing an applied force of +63
newtons.
part a: what is the force of friction acting on the box?
part b: determine the net force acting on the box.
part c: at what rate, if any, is the box accelerating?
formula list\tknown values\tunknown
$f_g = (m)(g)$
$f_n = - (f_g)$
$f_f = (\mu) (f_n)$
$f_{net} = f_a + f_f$
$f_{net} = (m)(a)$
$a = (f_{net}) / (m)$\t\t
work space

final answer with units

Explanation:

Step1: Find Force of Gravity ($F_g$)

Use $F_g = (m)(g)$. Mass $m = 25\,\text{kg}$, $g = 9.8\,\text{m/s}^2$.
$F_g = 25 \times 9.8 = 245\,\text{N}$.

Step2: Find Normal Force ($F_N$)

$F_N = -F_g$ (magnitude equal, opposite direction). So $F_N = 245\,\text{N}$ (magnitude).

Step3: Find Frictional Force ($F_f$)

Use $F_f = (\mu)(F_N)$. $\mu = 0.20$, $F_N = 245\,\text{N}$.
$F_f = 0.20 \times 245 = 49\,\text{N}$. (Friction opposes motion, so direction is -49 N if applied force is +63 N.)

Step4: Find Net Force ($F_{NET}$)

Use $F_{NET} = F_A + F_f$. $F_A = +63\,\text{N}$, $F_f = -49\,\text{N}$.
$F_{NET} = 63 - 49 = 14\,\text{N}$.

Step5: Find Acceleration ($a$)

Use $a = \frac{F_{NET}}{m}$. $F_{NET} = 14\,\text{N}$, $m = 25\,\text{kg}$.
$a = \frac{14}{25} = 0.56\,\text{m/s}^2$.

Part A Answer:

The force of friction is $\boldsymbol{49\,\text{Newtons}}$ (opposing the applied force).

Part B Answer:

The net force is $\boldsymbol{14\,\text{Newtons}}$.

Part C Answer:

The box accelerates at $\boldsymbol{0.56\,\text{m/s}^2}$.

(Note: For each part, follow the step-by-step. Above combines all steps for clarity. If answering each part separately, break down as per Step3 for A, Step4 for B, Step5 for C.)

Answer:

Step1: Find Force of Gravity ($F_g$)

Use $F_g = (m)(g)$. Mass $m = 25\,\text{kg}$, $g = 9.8\,\text{m/s}^2$.
$F_g = 25 \times 9.8 = 245\,\text{N}$.

Step2: Find Normal Force ($F_N$)

$F_N = -F_g$ (magnitude equal, opposite direction). So $F_N = 245\,\text{N}$ (magnitude).

Step3: Find Frictional Force ($F_f$)

Use $F_f = (\mu)(F_N)$. $\mu = 0.20$, $F_N = 245\,\text{N}$.
$F_f = 0.20 \times 245 = 49\,\text{N}$. (Friction opposes motion, so direction is -49 N if applied force is +63 N.)

Step4: Find Net Force ($F_{NET}$)

Use $F_{NET} = F_A + F_f$. $F_A = +63\,\text{N}$, $F_f = -49\,\text{N}$.
$F_{NET} = 63 - 49 = 14\,\text{N}$.

Step5: Find Acceleration ($a$)

Use $a = \frac{F_{NET}}{m}$. $F_{NET} = 14\,\text{N}$, $m = 25\,\text{kg}$.
$a = \frac{14}{25} = 0.56\,\text{m/s}^2$.

Part A Answer:

The force of friction is $\boldsymbol{49\,\text{Newtons}}$ (opposing the applied force).

Part B Answer:

The net force is $\boldsymbol{14\,\text{Newtons}}$.

Part C Answer:

The box accelerates at $\boldsymbol{0.56\,\text{m/s}^2}$.

(Note: For each part, follow the step-by-step. Above combines all steps for clarity. If answering each part separately, break down as per Step3 for A, Step4 for B, Step5 for C.)