Question 33a3c

Aug 26, 2016

Well.

Explanation:

There is only a downward force and no upward force so we will focus there.

$\sum {F}_{x}$ = $m \cdot g \cdot \sin \theta + 26.0 N - {f}_{k}$
$\sum {F}_{x}$ = $9 k g \cdot 9.8 \frac{m}{{s}^{2}} \cdot 0.54 + 26.0 N - \left[0.3 \cdot 9 k g \cdot 9.8 \frac{m}{{s}^{2}} \cdot 0.83\right]$
$\sum {F}_{x}$ = $47.6 + 26 N - 21.961 N$
$\sum {F}_{x}$ = $51.64 N$

Now, you are asked to find the velocity after t = 2 s and you know that the intial v is 0 since the box started from rest. You are going to have to use 1 of you kinematic equations

${v}_{f} = {v}_{o} + a \cdot t$

${v}_{o} = 0$
$t = 2 s$
v_f = ?
a = ?

How do you find acceleration? Well you have found the net downward Force so using Newton's 2nd law of motion

$F = m \cdot a$

$51.64 N$ = $9 k g$*$a$

$\frac{51.64 N}{9 k g}$ = $a$

$a$ = $5.73 \frac{m}{{s}^{2}}$

${v}_{f} = {v}_{o} + a \cdot t$
${v}_{f} = 0$ + $5.73 \frac{m}{{s}^{2}}$*$2 s$

${v}_{f} = 11.46 \frac{m}{s}$

Aug 27, 2016

$= 11.532 m {s}^{-} 1$

Explanation:

Question
A 9.00-kg box sits on a ramp that is inclined at 33.00 above the horizontal. The coefficient of friction between the box and the surface of the ramp is 0.300. A constant horizontal force F = 26.0 N is applied to the box (as in Figure given below) , and the box moves down the ramp.If the box is initially at rest, what is its speed 2.00 s after the force is applied? It is clear from the figure that the vertical component of applied force$F \sin \theta$ will diminish normal reaction but its horizontal component $F \cos \theta$increases the downward force.parallel to the plane of the ramp.
So
$\text{Normal Reaction } N = m g \cos \theta - F \sin \theta$

$\text{Frictional force } f = \mu N$

$\text{Net Downward force parallel to ramp}$

$= m g \sin \theta + F \cos \theta - \mu N$

$= m g \sin \theta + F \cos \theta - \mu \left(m g \cos \theta - F \sin \theta\right)$

$\text{Downward acceleration}$

$a = \frac{1}{m} \left(m g \sin \theta + F \cos \theta - \mu \left(m g \cos \theta - F \sin \theta\right)\right)$

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

Inserting
$m = 9 k g , g = 9.8 m {s}^{-} 2 , \mu = 0.3 , F = 26 N , \theta = {33}^{\circ}$

$\implies a = 9.8 \sin 33 + \frac{26}{9} \cos 33 - 0.3 \left(9.8 \cos 33 - \frac{26}{9} \sin 33\right)$

$\implies a = 5.337 + 2.423 - 0.3 \left(8.219 - 1.573\right) m {s}^{-} 2$

$\implies a = 5.337 + 2.423 - 1.994 m {s}^{-} 2 \approx 5.766 m {s}^{-} 2 \textcolor{red}{\text{ rounded up to 3 decimal place}}$

Now calculation of velocity 2s after the application of force F

${v}_{i} \to \text{Initial velocity} = 0$

$a \to \text{Acceleration} = 5.766 m {s}^{-} 2$

$t \to \text{Time} = 2 s$

v_f->"Final velocity"=?#

${v}_{f} = {v}_{i} + a \times t$

$\implies {v}_{f} = 0 + 5.766 \times 2 = 11.532 m {s}^{-} 1$