Question

when a 75 g mass is suspended from a vertical spring, the spring is stretched from a length of 4.0 cm to a length of 7.0 cm if the mass is pulled downward an additional 10 cm what is the total work done against the spring force.

Answer 1

i believe we have to figure out the constant of variation for the spring and determine a hookes law equation from it

Answer 2

w = kd ; w = 75, d = 3

k = 75/3 = 25

F(x) = 25x

\[\int_{7}^{7+10}25x\ dx\]

Answer 3

we integrate this to obtain:

12.5 x^2 evaluated from 7 to 17

12.5(17^2 - 7^2) should do it unless ive forgotten something

Answer 4

Whoops! amistre64 you forgot to multiply the mass by gravity and failed to observe the units! Otherwise, your methodology is correct.

Let's revisit to make sure we are clear on everything.

If we know the distance the spring stretches by attached some known mass to, we can determine the spring constant directly from Hook's Law (the knowledge of the spring constant is important to the second part of the question).

From Hook's Law we know that \[F_s = -kx\]Realizing that the force of the spring must equal to the weight of the mass, we can see that\[-mg = -kx \rightarrow mg = kx\]Keep in mind that \(x\) is the change in length, therefore, here \(x=0.03m\)We come up with the following expression\[(0.075 \cdot 9.81) [{\rm N}] = k \left [\rm N \over m \right] \cdot 0.03 [\rm m]\]

Once the spring constant is known, we can easily solve for work. Recalling first that\[F_s = -kx\] and second that\[W = \int\limits F dx\]Let's realize that the force we apply to the spring is of opposite sign of the spring force, \(F_s\), therefore for work we have\[W = \int\limits -F_x dx = kx dx = {1 \over 2}k x^2\]where \(x=0.1m\)

Answer 5

Equation for work is:
\[W = kx ^{2} /2\]
Given from the problem is x which is 10cm/ 0.1m
We have to find k.

\[F=-kx\]
So, solving for k:
k= (0.075 kg * 9.81) / (0.07-0.04)
k= 24.53

Going back to the 1st equation,
\[W= 24.53*0.1^2 / 2\]
W= 0.12.