Work-Energy Theorem Physics Example 4

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Example 4

hard
A 2 kg2 \text{ kg} block slides down a rough incline (μk=0.2\mu_k = 0.2) of height 5 m5 \text{ m} and length 10 m10 \text{ m}, starting from rest. What is its speed at the bottom? Use g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution

  1. 1
    Work done by gravity: Wg=mgh=2×9.8×5=98 JW_g = mgh = 2 \times 9.8 \times 5 = 98 \text{ J}.
  2. 2
    Normal force on incline: N=mgcos⁡θN = mg\cos\theta. With sin⁡θ=5/10=0.5\sin\theta = 5/10 = 0.5, cos⁡θ=0.866\cos\theta = 0.866. N=2(9.8)(0.866)=16.97 NN = 2(9.8)(0.866) = 16.97 \text{ N}.
  3. 3
    Work done by friction: Wf=−μkN⋅L=−0.2(16.97)(10)=−33.94 JW_f = -\mu_k N \cdot L = -0.2(16.97)(10) = -33.94 \text{ J}. Net work: W=98−33.94=64.06 JW = 98 - 33.94 = 64.06 \text{ J}. Speed: v=2Wm=128.122=64.06≈8.0 m/sv = \sqrt{\frac{2W}{m}} = \sqrt{\frac{128.12}{2}} = \sqrt{64.06} \approx 8.0 \text{ m/s}.

Answer

v≈8.0 m/sv \approx 8.0 \text{ m/s}
The work-energy theorem handles non-conservative forces elegantly. Gravity does positive work (adding KE), while friction does negative work (removing KE). The net work determines the final speed.

About Work-Energy Theorem

The net work done on an object by all forces acting on it equals the change in its kinetic energy.

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