A moving 4.80 kg block collides with a horizontal spring whose spring constant is 243 N/m. The...

Question:

A moving 4.80 kg block collides with a horizontal spring whose spring constant is 243 N/m. The block compresses the spring a maximum distance of 5.50 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.390. What is the work done by the spring in bringing the block to rest? How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? What is the speed of the block when it hits the spring?

Work-done by friction:

The work done by the friction is given by the product of the friction force and the displacement of the block. And the spring potential energy is directly proportional to the square of the velocity.

Answer and Explanation:

Given data:

  • Mass of the block {eq}\rm (m) = 4.8 \ kg {/eq}
  • Spring Constant {eq}\rm (k) = 243 \ N/m {/eq}
  • Maximum distance of compression {eq}\rm (x) = 5.5 \ cm {/eq}

Work done by the spring would be

{eq}\rm W = \dfrac{1}{2}kx^{2} \\ W = \dfrac{1}{2}(243) (0.055)^{2} \\ W = 0.367 \ J {/eq}

Now, the work done by the friction would be

{eq}\rm W_{f} = f \times x \\ W_{f} = (\mu\times mg)\times x \\ W_{f} = (0.39 \times 4.8 \times 9.8) \times (0.055) \\ W_{f} = 1.009 \ J {/eq}

Now, applying the conservation of energy

{eq}\begin{align} \rm \dfrac{1}{2}mv^{2} &= W + W_{f} \\ \rm \dfrac{1}{2}(4.8)(v^{2}) &= 0.367 + 1.009 \\ \rm v &= 0.757 \ m/s \\ \end{align} {/eq}


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