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SAT II Physics : Spring Force

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Example Question #1 : Forces

A 36kg $\displaystyle 36kg$ girl bounces on a massless pogo stick. If the spring constant for the stick is $15,000\frac{N}{m}$ $\displaystyle 15,000\frac{N}{m}$ , what is the maximum compression of the spring?

$g=-9.8\frac{m}{s^2}$ $\displaystyle g=-9.8\frac{m}{s^2}$

Possible Answers:

0.0004m $\displaystyle 0.0004m$

42.52m $\displaystyle 42.52m$

0.02m $\displaystyle 0.02m$

0.2m $\displaystyle 0.2m$

0.14m $\displaystyle 0.14m$

Correct answer:

0.02m $\displaystyle 0.02m$

Explanation:

There are two forces at work here: the force due to gravity and the restoring force of the spring. We can set these two forces equal to one another because the forces must be in equilibrium when the spring is compressed at its maximum point.

F_g=F_s $\displaystyle F_g=F_s$

Expand this equation by using the formulas for gravitational and spring force, respectively.

mg=-kx $\displaystyle mg=-kx$

Plug in our given values for the girl's mass, gravitational acceleration, and the spring constant. Using these values, we can solve for the displacement of the spring.

$(36kg)(-9.8\frac{m}{s^2})=-(15,000\frac{N}{m})(x)$ $\displaystyle (36kg)(-9.8\frac{m}{s^2})=-(15,000\frac{N}{m})(x)$

$-352.8N=-15,000\frac{N}{m}*x$ $\displaystyle -352.8N=-15,000\frac{N}{m}*x$

$\frac{-352.8N}{-15,000\frac{N}{m}}=x$ $\displaystyle \frac{-352.8N}{-15,000\frac{N}{m}}=x$

0.02m=x $\displaystyle 0.02m=x$

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Example Question #1 : Spring Force

A certain spring has an equilibrium length of 20cm $\displaystyle 20cm$ . When 20N $\displaystyle 20N$ of force are applied, the spring length becomes 30cm $\displaystyle 30cm$ . What is the spring constant?

Possible Answers:

$150\frac{N}{m}$ $\displaystyle 150\frac{N}{m}$

$100\frac{N}{m}$ $\displaystyle 100\frac{N}{m}$

$600\frac{N}{m}$ $\displaystyle 600\frac{N}{m}$

$400\frac{N}{m}$ $\displaystyle 400\frac{N}{m}$

$200\frac{N}{m}$ $\displaystyle 200\frac{N}{m}$

Correct answer:

$200\frac{N}{m}$ $\displaystyle 200\frac{N}{m}$

Explanation:

The equation for spring force is:

F_s=-kx $\displaystyle F_s=-kx$

The force applied to the spring will be equal and opposite the force of the spring itself. In this case, the force applied is 20N $\displaystyle 20N$ , so the force of the spring will be -20N $\displaystyle -20N$ .

$\displaystyle -20N=-kx$

The change in distance will be 10cm $\displaystyle 10cm$ , or 0.10m $\displaystyle 0.10m$ . Using this value, we can solve for the spring constant.

$\displaystyle -20N=-k(0.10m)$

$k=\frac{20N}{0.10m}=200\frac{N}{m}$ $\displaystyle k=\frac{20N}{0.10m}=200\frac{N}{m}$

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SAT II Physics : Spring Force

Study concepts, example questions & explanations for SAT II Physics

All SAT II Physics Resources

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Example Question #1 : Forces

Example Question #1 : Spring Force

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