Question

1. you are trying to slide a 50 kg box along the floor ($\mu = .8$). what force is needed to move your box along the floor at $3.1m/s^2$?
2. a box is sliding down the ramp at a constant velocity. if the force of friction is 16n and the angle of the ramp is $20^\circ$, what is the mass of the box?
3. a person is pulling a 30 kg box up a ramp that is at an angle of $11^\circ$. if the box is moving at $0.4m/s^2$, and the pulling force is 15n what is the coefficient of sliding friction?

Explanation:

(Problem 17):

Step1: Identify forces and Newton's law

The applied force $F$ must overcome friction and provide acceleration. Newton's second law: $F - f = ma$, where $f = \mu N = \mu mg$.

Step2: Rearrange for applied force

$F = ma + \mu mg = m(a + \mu g)$

Step3: Substitute values ($g=9.8m/s^2$)

$m=50kg$, $a=3.1m/s^2$, $\mu=0.8$
$F = 50\times(3.1 + 0.8\times9.8) = 50\times(3.1+7.84) = 50\times10.94$

(Problem 18):

Step1: Constant velocity means net force=0

The component of weight down the ramp equals friction: $mg\sin\theta = f$

Step2: Rearrange for mass $m$

$m = \frac{f}{g\sin\theta}$

Step3: Substitute values ($g=9.8m/s^2$)

$f=16N$, $\theta=20^\circ$
$m = \frac{16}{9.8\times\sin(20^\circ)} \approx \frac{16}{9.8\times0.3420}$

(Problem 19):

Step1: Apply Newton's second law up ramp

$F - mg\sin\theta - f = ma$, where $f = \mu N = \mu mg\cos\theta$

Step2: Rearrange for $\mu$

$\mu = \frac{F - m(a + g\sin\theta)}{mg\cos\theta}$

Step3: Substitute values ($g=9.8m/s^2$)

$F=15N$, $m=30kg$, $a=0.4m/s^2$, $\theta=11^\circ$
$\mu = \frac{15 - 30\times(0.4 + 9.8\times\sin(11^\circ))}{30\times9.8\times\cos(11^\circ)} \approx \frac{15 - 30\times(0.4+1.86)}{30\times9.8\times0.9816}$

Answer:

1. Problem 17: $547\ N$
2. Problem 18: $\approx 4.7\ kg$
3. Problem 19: $\approx 0.02$