To stretch a spring 8,5 cm from its unstretched length, 251 J of work must be done. What is the stifness constant (spring constant) of this spring in N/m ? (in N) Round off your result to 1 decimal place.

To find the spring constant (k), we can use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched or compressed. The formula for Hooke's Law is:<br /><br />\[ F = k \cdot x \]<br /><br />where:<br />- \( F \) is the force applied to the spring,<br />- \( k \) is the spring constant,<br />- \( x \) is the distance the spring is stretched or compressed from its unstretched length.<br /><br />Given that 251 J of work must be done to stretch the spring 8.5 cm (0.085 m), we can use the work done on the spring to find the force. The work done on a spring is given by:<br /><br />\[ W = \frac{1}{2} k x^2 \]<br /><br />where:<br />- \( W \) is the work done on the spring,<br />- \( k \) is the spring constant,<br />- \( x \) is the distance the spring is stretched.<br /><br />We can rearrange this formula to solve for the spring constant \( k \):<br /><br />\[ k = \frac{2W}{x^2} \]<br /><br />Substituting the given values:<br /><br />\[ k = \frac{2 \times 251 \, \text{J}}{(0.085 \, \text{m})^2} \]<br /><br />\[ k = \frac{502 \, \text{J}}{0.007225 \, \text{m}^2} \]<br /><br />\[ k = 69400 \, \text{N/m} \]<br /><br />Rounding off to 1 decimal place, the spring constant \( k \) is:<br /><br />\[ k = 69400 \, \text{N/m} \]