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Nookie1986 [14]

3 years ago

15

A force of 36.0 N is required to start a 3.0-kg box moving across a horizontal concrete floor. Part A) What is the coefficient o

f static friction between the box and the floor? Express your answer using two significant figures. Part B) If the 36.0-N force continues, the box accelerates at 0.54 m/s^2. What is the coefficient of kinetic friction? Express your answer using two significant figures.

1 answer:

matrenka [14]3 years ago

4 0

Answer:

A) 1.2

B) 1.1

Explanation:

A)

F = force required to start the box moving = 36.0 N

m = mass of the box = 3 kg

$F_{g}$ = weight of the box = mg = 3 x 9.8 = 29.4 N

$F_{n}$ = normal force acting on the box by the floor

normal force acting on the box by the floor is given as

$F_{n}$ = $F_{g}$ = 29.4

$F_{s}$ = Static frictional force = F = 36.0 N

$\mu _{s}$ = Coefficient of static friction

Static frictional force is given as

$F_{s}$ = $\mu _{s}$ $F_{n}$

36.0 = $\mu _{s}$ (29.4)

$\mu _{s}$ = 1.2

B)

a = acceleration of the box = 0.54 m/s²

F = force applied = 36.0 N

$f_{k}$ = kinetic frictional force

$\mu _{k}$ = Coefficient of kinetic friction

force equation for the motion of the box is given as

F - $f_{k}$ = ma

36.0 - $f_{k}$ = (3) (0.54)

$f_{k}$ = 34.38 N

Coefficient of kinetic friction is given as

$\mu _{k}=\frac{f_{k}}{F_{n}}$

$\mu _{k}=\frac{34.38}{29.4}$

$\mu _{k}$ = 1.1

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represents the velocity, again the derivative of velocity gives us acceleration

So

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now to the time when the rope made angle of 30° with the water

generally angular velocity is mathematically represented as

$w = \frac{\Delta \theta}{\Delta t}$

Where $\theta$ is the angular displacement

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$2 = \frac{30 -20 }{t -0}$

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$\alpha_t = r \theta'' + 2 r' \theta '$

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So

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

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The net resultant acceleration is mathematically represented as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the direction of the is acceleration is mathematically represented as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

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$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

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$F_{net}= 1.837 *10^4N$

The direction of the force is

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

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