Question

suppose that 2 j of work is needed to stretch a spring from its natural length of 32 cm to a length of 43 cm. (a) how much work is needed to stretch the spring from 37 cm to 41 cm? (round your answer to two decimal places.) j (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:

Step1: Calculate the spring - constant (k)

First, use Hooke's law $F = kx$. When $F = 25N$, we need to find $k$. But we first consider the work - energy relationship for part (a). The work done on a spring is given by $W=\frac{1}{2}k(x_f^2 - x_i^2)$. The initial length of the spring is $x_i = 32cm=0.32m$ and the final length is $x_f = 43cm = 0.43m$. The work $W = 2J$. We know that $W=\frac{1}{2}k(x_f^2 - x_i^2)$. So, $2=\frac{1}{2}k((0.43)^2-(0.32)^2)$. First, calculate $(0.43)^2-(0.32)^2=0.43^2 - 0.32^2=0.1849 - 0.1024 = 0.0825$. Then, from $2=\frac{1}{2}k\times0.0825$, we can solve for $k$: $k=\frac{2\times2}{0.0825}=\frac{4}{0.0825}\approx48.5N/m$.

Step2: Calculate the work for part (a)

The spring is stretched from $x_i = 37cm = 0.37m$ to $x_f = 41cm=0.41m$. Using the work - formula $W=\frac{1}{2}k(x_f^2 - x_i^2)$, substitute $k\approx48.5N/m$, $x_i = 0.37m$ and $x_f = 0.41m$. First, calculate $x_f^2 - x_i^2=(0.41)^2-(0.37)^2=0.1681 - 0.1369 = 0.0312$. Then $W=\frac{1}{2}\times48.5\times0.0312 = 0.75J$.

Step3: Calculate the displacement for part (b)

Using Hooke's law $F = kx$, where $F = 25N$ and $k\approx48.5N/m$. Then $x=\frac{F}{k}=\frac{25}{48.5}\approx0.515m = 51.5cm$.

Answer:

(a) $0.75J$
(b) $51.5cm$