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Shkiper50 [21]
1 year ago
15

4 Four objects are stiuated along the y axis as follow: a 20 kg object is at +3 m, a 3 kg object is at +2.5 m, a 2.5 kg object i

s at origin, and a 4 kg object is at -5 m. What is the center of mass of these objects?​
Physics
1 answer:
FinnZ [79.3K]1 year ago
4 0

Answer: 1.348 meters

Explanation: Although the sign is missing from the location of the 4.00 kg object, it is assumed to be positive. The net moment of all the objects about the center of mass must be zero. Let the center of mass be on the y axis at a point  c . Adding the four moments together, we get:

(2.00)(3.00−c)+(3.00)(2.50−c)+(2.50)(0−c)+(4.00)(0.500−c)=0

6.00−2.00c+7.50−3.00c+0−2.50c+2.00−4.00c=0

11.5c=15.50

c=  1.348 metres

The center of mass is on the y axis at  y  = 1.348 metres.

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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

           y is the difference in weight which is   =1.10-0.50=0.6m

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}

                       = 8.17 N/m

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

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

5 0
3 years ago
Read 2 more answers
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