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.

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

Vertical forces:

[tex]\sum F_{y}=W-F_{p_{y} } -F_{\mu} =ma_{y}[/tex]

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

Then,

[tex]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})[/tex]

Now, we solve for [tex]a_{y}[/tex]

[tex]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}[/tex]

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

[tex]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[/tex]

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

In this case, the sum of vertical forces would be

[tex]\sum F_{y}=W-F_{p_{y} } -F_{\mu} =0[/tex]

Where [tex]F_{p_{y} }= F_{p}sin(45\°)[/tex]

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

[tex]30 - F_{p}sin(45\°)-0.516 F_{p}=0[/tex]

Then, we solve for [tex]F_{p}[/tex]

[tex]30-0.71F_{p}-0.516 F_{p}=0\\30=1.226 F_{p}\\ F_{p}=\frac{30}{1.226} \approx 24.5 \ N[/tex]

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

okpalawalter8 okpalawalter8 · Answer 2 · 2021-09-21T18:57:43+02:00

We have that the speed of the box 2.0s after you started pushing on it is

v=3m/s

From the question we are told that

A box of mass 3.1kg slides down a rough vertical wall.

he gravitational force on the box is 30N

the box reaches a speed of 2.5m/s

a constant force of magnitude Fp = 23N

the box and the wall of magnitude 13N

2.0s after you started pushing on it

the magnitude of the friction force is f=0.516Fp.

Generally the equation for Component force is mathematically given as

[tex]Fv_p=Fpsin\theta\\\\Fv_p=Fpsin45\\\\Fv_p=23{\frac{1}{\sqrt{2}}}[/tex]

[tex]Fv_p=16.26N[/tex]

Generally the Newtons equation for Force is mathematically given as

[tex]F_g-F_f-F_p=ma\\\\Therefore\\\\30-13-16.3N=(-0.736)a\\\\a=0.23m/s^2\\\\[/tex]

Generally the Newtons equation for Motion is mathematically given as

[tex]v=u+at\\\\Therefore\\\\\v=(2.5)+0.23*2\\\\\v=3m/s[/tex]

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