Question

A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed of 2.5m/s , you start pushing on one edge of the box at a 45° angle (use degrees in your calculations throughout this problem) with a constant force of magnitude Fp = 23N . There is now a frictional force between the box and the wall of magnitude 13N . How fast is the box sliding 2.0s after you started pushing on it?
Assuming that the angle at which you push on the edge of the box is again 45∘, with what magnitude of force Fp should you push if the box were to slide down the wall at a constant velocity? Note that, in general, the magnitude of the friction force will change if you change the magnitude of the pushing force. Thus, for this part, assume that the magnitude of the friction force is f=0.516Fp.

Answer 1

Answer:

Explanation:

a) The net force on the box in vertical direction:

Fnet=Fg−f−Fp  *sin45  °

 here Fg is the  gravitational force .f is the force of friction and , Fp is the pushing force.

Fnet=ma

ma=Fg−f−Fp  *sin45  °

​a=$\frac{30-13-23*sin(45)}{3.1}$

 =0.24 m/s²  

Vf =Vi +at

     =0.48+0.24*2

Vf=2.98 m/s

b)

                                Fnet=Fg−f−Fp  *sin45  °

                                        =Fg−0.516Fp−Fp  *sin45  °

                                        =30-1.273Fp

                                Fnet=0               (Because speed is constant)

                                Fp=30/1.273

                                      =23.56 N

Answer 2

Answer:

$v_{f} \approx 2.1 \ m/s$ (after two seconds)

$F_{p} \approx 24.5 \ N$ (with no acceleration)

Explanation:

Givens

$m=3.14 \ kg$

$W=30 \ N$

$v_{0} = 2.5 \ m/s$

$\theta = 45\°$

$F_{p}=23 \ N$ (constant)

$F_{\mu}=13 \ N$ (friction)

$t= 2.0 \ sec$

First, we have to find the speed after 2 seconds, when its initial speed is $v_{0} = 2.5 \ m/s$. The acceleration can be found by using Newton's Second Law

Vertical forces:

$\sum F_{y}=W-F_{p_{y} } -F_{\mu} =ma_{y}$

Where $F_{p_{y} }= F_{p}sin(45\°)$, which is deducted from the right angle formed (refer to the image attached).

Then,

$W - F_{p}sin(45\°) - F_{\mu} = m a_{y}\\ 30 \ N - (23 \ N)\frac{\sqrt{2} }{2}-13 \ N =3.14 \ kg (a_{y})$

Now, we solve for  $a_{y}$

$30-\frac{23\sqrt{2} }{2}-13=3.14 a_{y}\\ 17 - \frac{23\sqrt{2} }{2}=3.14 a_{y}\\\frac{34-23\sqrt{2}}{2}= 3.14 a_{y}\\a_{y}= \frac{34-23\sqrt{2}}{6.28} \approx 0.2 \ m/s^{2}$

Then, we use the following formula to find the speed after 2 seconds

$v_{f}=v_{0}-at \\ v_{f}=2.5 \ m/s - (0.2 \ m/s^{2})(2sec)=2.5- 0.4=2.1 \ m/s$

Now, if the friction force is $F_{\mu}=0.516 F_{p}$ and the box falls with constant speed, what would be the magnitude of $F_{p}$ to happen.

In this case, the sum of vertical forces would be

$\sum F_{y}=W-F_{p_{y} } -F_{\mu} =0$

Where $F_{p_{y} }= F_{p}sin(45\°)$

Notice that the net vertical force is zero, because there's no acceleratio, the box is moving at a constant speed.

$30 - F_{p}sin(45\°)-0.516 F_{p}=0$

Then, we solve for $F_{p}$

$30-0.71F_{p}-0.516 F_{p}=0\\30=1.226 F_{p}\\ F_{p}=\frac{30}{1.226} \approx 24.5 \ N$

Therefore, the force needed to have no acceleration is 24.5 N.