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Montano1993 [528]
3 years ago
9

If Superman wants to slow down a fast car with speed 20 m/s and mass 1000 kg, how much force in N does he need to apply if he wa

nts to get it down to zero within 3.2 seconds?
Physics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

Force of 6250N is required to slow down the body to zero.

Explanation:

Speed(V)=20m/s

Initial speed (u) =0m/s

Mass(M)=1000kg

Time(t)=3.2seconds

Using first equation of motion to find the acceleration of the body:

V=U + at

20m/s=0 + a*3.2s

Therefore, a=(20m/s)/3.2s

a=6.25m/s^2

To find the force required to stop the body or get it down to zero:

Force(F)=mass x acceleration

F=(1000kg)*(6.25m/s^2)

F=6250N.

Therefore a force of 6250N is required to slow it down.

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s = 6.25 10⁻²² m

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A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
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The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

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To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

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F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which  2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t)  we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and  u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4)

The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

8 0
4 years ago
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