4. +-0.83 points SCalcET8 6.4.501.XP My Notes A spring has a natural length of 2

Question

4. +-0.83 points SCalcET8 6.4.501.XP My Notes A spring has a natural length of 24 cm. If a 26-N force is required to keep it stretched to a length of 30 cm, how much work W is required to stretch it from 24 cm to 27 cm? Round your answer to two decimal places.) Submit Answer 5, ÷ 0.83 points SCalcET8 6.4.010 My Notes If the work required to stretch a spring 2 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch it 6 in. beyond its natural length? ft-lb 6. +-0.83 points SCalcET8 6.4.009. My Notes Suppose that 3 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 40 cm. (a) How much work is needed to stretch the spring from 30 cm to 35 cm? (Round your answer to two decimal places.) (b) How far beyond its natural length will a force of 25 N keep the spring stretched? (Round your answer one decimal place.) cm

Explanation / Answer

Solution 4:

Assuming the spring obeys Hooke's Law, we see that the force required to stretch the spring x units beyond its natural length is:
F = kx, for some constant of proportionality k.

Since 26 N is required to stretch the spring 0.06m beyond its natural length (the natural length of the spring is 24 cm and we stretch it to 30 cm, so this is 6 cm
beyond the natural length;
also, we want to convert the distance to meters so we can get the required units in the end), we see that the value of k in this case is:
26 N = k(0.06 m) ==> k = 26/0.06 = 433.3 N/m.
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So, the force required to stretch the spring x units beyond its natural length is:
F = 433.3x N (with x in meters).

Integrating this from x = 0 (the natural length) to x = 0.03 (from the 24->27 cm stretch) gives the required work to be:
W = 433.3x dx (from x=0 to 0.03) = 433.3(x^2)/2 (from x=0 to 0.03) = 0.194J

Answer = 0.194J

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