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Kamila [148]
4 years ago
6

23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n

orth with a magnitude of 130.0 N and Jennifer’s force is 32° east of north with a magnitude of 180.0 N. (a) Find the net force in component form. (b) Find the magnitude and direction of the net force. (c) If Andrea and Jennifer’s housemates, David and Stephanie, disagree with the move and want to prevent its relocation, with what combined force F → DS should they push so that the couch does not move?
Physics
1 answer:
PSYCHO15rus [73]4 years ago
4 0

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

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m= mass of the particle moving upward

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h= distance above the the surface

Part a

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The initial kinetic energy would be \frac{1}{2}mv^2. The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

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Dividing both sides by m and replacing \frac{GM}{R} by a_g R

And the equation (1) becomes:

-a_g R+\frac{1}{2} v^2=0   (2)

If we solve for v we got this:

v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s

Part b

When we consider a particule at this surface at the starting point we have that:

U_i=-\frac{GMm}{R}

K_i=\frac{1}{2}mv^2

Considering that the particle is at a distance h above the surface and then stops we have that:

U_f=-\frac{GMm}{R+h}

K_f=0

And the balance of energy would be:

-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}

Dividing again both sides by m and replacing \frac{GM}{R} by a_g R^2 we got:

-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}

If we solve for h we can follow the following steps:

R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

h=\frac{2a_g R^2}{2a_g R-v^2}-R

Replacing:

h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

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And after the particle reach the asteroid we have this:

U_f=-\frac{GMm}{R}

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So the balance of energy would be:

-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2

Replacing again a_g R^2 instead of GM and dividing both sides by m we have:

-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2

And solving for v:

a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2

Multiplying both sides by two and taking square root:

v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}

Replacing

v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s

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3 years ago
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