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Novay_Z [31]
3 years ago
5

Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is halved, then what is the ne

w acceleration of the sled?
(show how you solved it)
Physics
1 answer:
yanalaym [24]3 years ago
3 0

Start with Newton's 2nd law of motion:

                             Force = (mass) x (acceleration) .

Since we're going to be talking about acceleration,
lets divide each side by (mass):

                             Acceleration = (force) / (mass) .

OK.  We start out with a certain acceleration 'A₀'.  It's the result of
a certain force 'F₀' and a certain mass 'M₀'.  (I used the little subscripts " ₀ "
to show that these are the originals, before any changes.)

             Original acceleration = (Original force) / (original mass)

                              A₀             =          F₀            /        M₀ .

Now you want to triple the force and cut the mass in half: 

                     New acceleration  A₁ = (3 F₀) / (1/2 M₀) .

Divide each side by 3:           A₁ / 3  =  F₀ / (1/2 M₀) .

Multiply each side by  1/2 :   (1/2 A₁) / 3  =  F₀/M₀

                                                   A₁ / 6  =  F₀/M₀

Take a look at the right side of that equation . . .  F₀/M₀ .
That's just the original acceleration  A₀ .
So now, after the change, we have                 A₁ / 6  =  A₀ .

You asked "What is the new acceleration ?"
OK.  Multiply each side by  6  :                        A₁ = 6 A₀ .

Whatever the original acceleration was, the
new acceleration is 6 times as much.

If it was originally  2 m/s², then after the change, it becomes  (6 x 2)  =  12 m/s² .

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The equation that relates distance, velocities, acceleration, and time is,
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The value of t from this equation is 24.73 s

(2) Thrown rock with V₀ = 26 m/s
                (3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
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The difference between the tim,
        difference = 24.73 s - 5.61 s
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<em>ANSWER: 19.12 s</em>
5 0
3 years ago
Read 2 more answers
A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the
ser-zykov [4K]

Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

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The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

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The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

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So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

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