A spring of constant \(k=340 \mathrm{N} / \mathrm{m}\) is used to launch a \(1.5-\mathrm{kg}\) block along a horizontal surface whose coefficient of sliding friction is \(0.27 .\) If the spring is compressed \(18 \mathrm{cm},\) how far does the block slide?

Short Answer

Expert verified
The block slides approximately 1.40 m.

Step by step solution

01

Calculate the Initial Potential Energy of the Spring

The potential energy stored in a compressed or stretched spring is given by the formula \(E_P = \frac{1}{2} k x^2\). Substituting in our values gives: \(E_P = \frac{1}{2} \times 340 \, \mathrm{N/m} \times (0.18 \, \mathrm{m})^2 = 5.508 \, \mathrm{J}\).
02

Find the Work Done Against Friction

The work done against friction is equal to the frictional force multiplied by the distance the block slides, or \(W = \mu m g d\), where \(g\) is the acceleration due to gravity (9.8 m/s²) and \(d\) is the distance we're solving for.
03

Apply the Principle of Energy Conservation

The work done against friction is equal to the initial potential energy stored in the spring (due to energy conservation), so we solve the equation \(5.508 \, \mathrm{J} = \mu m g d\) for \(d\). Substituting our known values, we get \(d = \frac{5.508 \, \mathrm{J}}{0.27 \times 1.5 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2}} \approx 1.40 \, \mathrm{m}\).

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