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Ackbach
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Problem: A light horizontal spring has a spring constant of $105 \; \text{N/m}$. A $2.00\; \text{kg}$ block is pressed against one end of the spring, compressing the spring $0.100 \; \text{m}$. After the block is released, the block moves $0.250 \; \text{m}$ to the right before coming to rest. What is the coefficient of kinetic friction between the horizontal surface and the block?
My attempt: Since there is friction, Conservation of Mechanical Energy does not apply. The Work-Energy Theorem does, however: $W_{ \text{net}}= \Delta KE$. Let us label the compressed state 1, the natural length of the spring state as 2, and the stopped state as 3. Indices will correspond to these states. If we examine the transition from 1 to 2, then
$$W_{ \text{net}}=KE_{f}= \frac{1}{2} mv_{2}^{2}.$$
But $W_{ \text{net}}=W_{el}-W_{f_{12}}$, so
$$W_{el}= \mu_{k}mgx_{12}+ \frac{1}{2} mv_{2}^{2}.$$
Here $W_{f_{12}}$ represents the work friction does in moving from 1 to 2, and $x_{12}$ is the distance from 1 to 2.
If we examine the transition from 2 to 3, we have that
$$-W_{f_{23}}=- \frac{1}{2} mv_{2}^{2} \quad \implies \quad
\mu_{k} mgx_{23}= \frac{1}{2} mv_{2}^{2}.$$
Therefore,
$$W_{el}= \mu_{k}mgx_{12}+ \mu_{k}mgx_{23}= \mu_{k}mgx_{13}.$$
Hence,
$$ \mu_{k}= \frac{W_{el}}{mgx_{13}}.$$
Now $W_{el}$ is an unknown. The question is, how can I compute $W_{el}$ without resorting to
$$W_{el}=\int \vec{F}_{el} \cdot d \vec{r} \; ?$$
If you compute this integral, you'll see that
$$W_{el}= \frac{1}{2} kx_{12}^{2}.$$
And if I carry out the rest of the algebra, I get
$$ \mu_{k}= \frac{k x_{12}^{2}}{2mgx_{13}}=
\frac{ (105 \; \text{N/m}) (0.1 \; \text{m})^{2}}{2 (2.0 \; \text{kg})
(9.8 \; \text{m/s}^{2}) (0.25 \; \text{m})} = 0.107,$$
which agrees with the book's answer.
I was thinking you might be able to do $W_{el}=- \Delta PE$, but doesn't that depend on the Conservation of Mechanical Energy? The other issue with using $W_{el}=- \Delta PE$ is that Holt Physics does not contain this equation.
So, just to be clear: my question is, "How can you solve this problem without using calculus?" It appears to me that friction is operational when the spring is pushing the block, so you can't just use conservation of mechanical energy for 1 to 2, and continue on from there.
My attempt: Since there is friction, Conservation of Mechanical Energy does not apply. The Work-Energy Theorem does, however: $W_{ \text{net}}= \Delta KE$. Let us label the compressed state 1, the natural length of the spring state as 2, and the stopped state as 3. Indices will correspond to these states. If we examine the transition from 1 to 2, then
$$W_{ \text{net}}=KE_{f}= \frac{1}{2} mv_{2}^{2}.$$
But $W_{ \text{net}}=W_{el}-W_{f_{12}}$, so
$$W_{el}= \mu_{k}mgx_{12}+ \frac{1}{2} mv_{2}^{2}.$$
Here $W_{f_{12}}$ represents the work friction does in moving from 1 to 2, and $x_{12}$ is the distance from 1 to 2.
If we examine the transition from 2 to 3, we have that
$$-W_{f_{23}}=- \frac{1}{2} mv_{2}^{2} \quad \implies \quad
\mu_{k} mgx_{23}= \frac{1}{2} mv_{2}^{2}.$$
Therefore,
$$W_{el}= \mu_{k}mgx_{12}+ \mu_{k}mgx_{23}= \mu_{k}mgx_{13}.$$
Hence,
$$ \mu_{k}= \frac{W_{el}}{mgx_{13}}.$$
Now $W_{el}$ is an unknown. The question is, how can I compute $W_{el}$ without resorting to
$$W_{el}=\int \vec{F}_{el} \cdot d \vec{r} \; ?$$
If you compute this integral, you'll see that
$$W_{el}= \frac{1}{2} kx_{12}^{2}.$$
And if I carry out the rest of the algebra, I get
$$ \mu_{k}= \frac{k x_{12}^{2}}{2mgx_{13}}=
\frac{ (105 \; \text{N/m}) (0.1 \; \text{m})^{2}}{2 (2.0 \; \text{kg})
(9.8 \; \text{m/s}^{2}) (0.25 \; \text{m})} = 0.107,$$
which agrees with the book's answer.
I was thinking you might be able to do $W_{el}=- \Delta PE$, but doesn't that depend on the Conservation of Mechanical Energy? The other issue with using $W_{el}=- \Delta PE$ is that Holt Physics does not contain this equation.
So, just to be clear: my question is, "How can you solve this problem without using calculus?" It appears to me that friction is operational when the spring is pushing the block, so you can't just use conservation of mechanical energy for 1 to 2, and continue on from there.
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