A 250 g woodblock firmly attached to a horizontal spring slides along a table with a coefficient...

Question:

A 250 g woodblock firmly attached to a horizontal spring slides along a table with a coefficient of friction of 0.40. A force of 30.N compresses the spring 20 cm. If the spring is released from this position, what will be the block's speed when it passes equilibrium position?

Conservative and Non-conservative force:

The work done by the force is independent of the path, then the force is known to be conservative, and if the work-done depends upon the path, then the force must be non-conservative.

Answer and Explanation:

Given data

  • Force applied {eq}\rm (F) = 30 \ N {/eq}
  • Displacement {eq}\rm (x) = 0.2 \ cm {/eq}
  • Coefficient of friction {eq}\rm (\mu) = 0.4 {/eq}

Now, the energy that should be stored in the spring due to the force

{eq}\rm W = F \times x \\ W = 30 \times 0.02 \\ W = 0.6 \ J {/eq}

Now, the work done by the friction would be

{eq}\rm W_{f} = 2(f \times x ) \\ W_{f} = 2 \times \mu mg \times x \\ W_{f} = 2\times 0.4 \times 0.25 \times 9.8 \times 0.02 \\ W_{f} = 0.392 \ J {/eq}

Now, the kinetic energy of the block would be

{eq}\begin{align} \rm K &= \rm W - W_{f} \\ \rm 0.5 mv^{2} &= \rm W - W_{f} \\ \rm 0.5 \times 0.25 \times v^{2} &= 0.6 - 0.392 \\ \rm v &= 1.29 \ m/s \\ \end{align} {/eq}


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