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anzhelika [568]
3 years ago
10

Your friend provides a solution to the following problem. Evaluate her solution. The problem: Jim (mass 50 kg) steps off a ledge

that is 2.0 m above a platform that sits on top of a relaxed spring of force constant 8000 N/m. How far will the spring compress while stopping Jim? Your friend's solution: (50kg)(9.8m/s2)(2.0m)=(1/2)(8000N/m)x x=0.25m4.
Physics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

No its wrong

Correct compression is 0.41

Explanation:

After jumping from 2m height on spring attached platform platform is compressed a distance x(let).

So work done by gravity on Jim is converted into spring potential energy.

k=8000 N/m

mass of Jim =50 kg

\frac{1}{2}k(x)^{2} =m*g(2+x)

\frac{1}{2}8000(x)^{2} =50*9.8(2+x)

8.16x^{2} =2 +x

Solve this quadratic one solution is positive and other is negative.

positive one is our answer = 0.41 m

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g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe
olga_2 [115]

Answer:

The minimum compression is  x= 0.046m

Explanation:

From the question we are told that

              The mass of the block is m_b = 1.45 kg

               The spring constant is  k = 860 N/m

               The coefficient of static friction is  \mu = 0.36

For the the block not slip it mean the sum of forces acting on the  horizontal axis is equal to the forces acting on the vertical axis

     Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

                   F_y = m_b*g

And the force acting on the horizontal axis is  force due to the spring which is mathematically represented as

                   F_x = k *x * \mu

where x is the minimum compression to keep the block from slipping

        Now equating this two formulas and making x the subject

                      x = \frac{m_b * g}{k * \mu}

substituting values we have

                     x = \frac{1.45 * 9.8}{860 *0.36}

                        x= 0.046m

 

3 0
3 years ago
A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the
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K.E =  \frac{1}{2} m {v}^{2}  \\  {v}^{2}_i  =  {v}^{2} _x + {v}^{2} _y \\  =  {(6)}^{2}  +  {( - 2)}^{2}  = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (40) = 60J \\  \\ {v}^{2}_f  =  {v}^{2} _x + {v}^{2} _y \\  =  {(8)}^{2}  +  {(4)}^{2}  = 80m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\  = 120J - 60J \\  = 60J

3 0
2 years ago
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