Example 6.5 Measuring k for a Spring A common technique used to measure the forc

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Example 6.5 Measuring k for a Spring A common technique used to measure the force constant of a spring is demonstrated by the setup in the figure. The spring is hung vertically (figure (a)), and an object of mass m is attached to its lower end. Under the action of the "load" mg, the spring stretches a distance d from its equilibrium position (figure (b) (A) If a spring is stretched 1.9 cm by a suspended object having a mass of 0.40 kg, what is the force constant of the spring? (B) How much work is done by the spring on the object as it stretches through this distance? veng SOLVE IT Conceptualize Consider figure (b), which shows what happens to the spring when the object is attached to it. Simulate this situation by hanging an object on a rubber band Determining the force constant k of a spring. The elongation d is caused by the weight mg of the attached object. Categorize The object in figure (b) is not accelerating, so it is modeled as a particle in equilibrium Analyze Because the object is in equilibrium, the net force on it is zero and the gravitational force m (figure (c)) upward spring force balances the downward (A) If a spring is stretched 1.9 cm by a suspended object having a mass of 0.40 kg, what is the force constant of the spring? Apply the particle in equilibrium model to the object: kd and Apply Hooke's law to give Fs solve for k k mo 0.40 kg)19.80 m/s) 1.9 × 10-2 m N/m (B) How much work is done by the spring on the object as it stretches through this distance? Use the equation to find the work done by the spring on the object -- (k)(1.9 x 10 m) Finalize As the object moves through the 1.9 cm distance, the gravitational force also does work on it. This work is positive because the gravitational force is downward and so is the displacement of the point of application of this force. Based on the equation and the discussion afterward, would we expect the work done by the gravitational force to be the work done by the spring? Let's find out. Evaluate the work done by the gravitational force on the object: w = F·M·= (mg)(d) cos 0 = mgd = (0.40 kg)(9.80m/s*)(1.9 × 10-2 m) If you expected the work done by gravity simply to be that done by the spring with a positive sign, you may be surprised by this result! To understand why that is not the case, we need to explore further, as we do in the next section. MASTER IT HINTS: GETTING STARTED I M STUCK A spring with spring constant k = 590 N/m is initially compressed x = 1.70 cm. It is now allowed to relax back to its equilibrium length and then it is stretched an additional 2.50 cm. How much work is done by the spring during this process

Explanation / Answer

A) for A

k = 0.40 * 9.80/0.019

k = 206 N/m

B)

for the work done

W = - 0.50 * 206 * 0.019^2

W = - 0.0372 J

Finalize

W = m *g * d

W = 0.40 * 9.8 * 0.019

W = 0.0745 J

master it

work done by spring = 0.50 * 590 * ((0.025 + 0.017)^2 - 0.017^2)

work done by spring = 0.435 J

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