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5.(20 points) A block with mass m_1=5kg is attached to a block with mass m_2=8kg by a massless string that passes over a massless pulley as shown. The coefficient of friction between that table and block 1 is mu_k=0.2. B

Concept: Work-Energy Theorem

Question

5.(20 points) A block with mass m_{1}=5kg is attached to a block with mass m_{2}=8kg by
a massless string that passes over a massless pulley as shown. The coefficient of friction
between that table and block 1 is μ\mu_{k}=0.2. Block 1 is attached to a spring with spring
constant k=140N/m. The system is release from rest when the string is unstretched.
Determine the speed of the two blocks when they have moved 0.5m.
Wnet=ΔKEW_{net} = \Delta KE

Try it yourself first

Hint: Think about all the forces acting on the system and how they contribute to the work done as the blocks move.

Step-by-step solution

  1. 1

    Work by Gravity

    Wg=m2∗g∗dW_g = m2 * g * d
    Wg=m2gdW_g = m_2gd
  2. 2

    Work by Friction

    W_f = -μk * N * d • N = m1 * g
    Wf=−μkm1gdW_f = -\mu_k m_1gd
  3. 3

    Work by Spring

    Ws=−0.5∗k∗d2W_s = -0.5 * k * d^2
    Ws=−12kd2W_s = -\frac{1}{2}kd^2
  4. 4

    Net Work & KE

    W_net = W_g + W_f + W_s • ΔKE = 0.5 * (m1+m2) * v^2
    Wnet=12(m1+m2)v2W_{net} = \frac{1}{2}(m_1+m_2)v^2
  5. 5

    Solve for Speed

    v = 2∗Wnet/(m1+m2)\sqrt{2 * W_net / (m1+m2)}
    v=2Wnetm1+m2v = \sqrt{\frac{2W_{net}}{m_1+m_2}}

Final answer

1.61 m/s

Remember: Use work-energy theorem when forces are variable or distance is given.

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When a spring is compressed, the work done BY the spring is:

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