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leva [86]
3 years ago
11

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.

Physics
2 answers:
Nostrana [21]3 years ago
5 0

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

stepan [7]3 years ago
5 0

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.

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The initial potential energy of the wagon containing gold boxes will enable

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Location of the rangers = 41 meters from the canyon

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Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

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Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

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v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

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Momentum = Mass, m × Velocity, v

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By conservation of momentum, we have;

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Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

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snow_lady [41]

Answer:

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