College Physics
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![Below is a conservation of energy problem. The solution to this problem is provided. Assess whether the solution provided is correct or incorrect AND EXPLAIN WHY.
A 4.6 kg block is at rest against a horizontal spring that is compressed by 0.40 m. The spring has a spring constant of \( k_1 = 2750 \, \text{N/m} \). After leaving the spring, it travels up a 28° incline to a height of 2.8 m. At the top of the hill is a second spring with a spring constant of \( k_2 = 350 \, \text{N/m} \). The horizontal portions are frictionless, but the hill has a coefficient of kinetic friction equal to \( u_k = 0.16 \). The final velocity of the block is \( 9.78 \, \text{m/s} \).
**How much work is done by friction as the block slides up the hill?**
\[ W = F \cos \theta d \]
\[ F = F_k = u_k N \]
\[ u_k = 0.16 \]
\[ N = mg = (4.6 \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 45.1 \, \text{N} \]
\[ \theta = 28° \]
\[ d = 2.80 \, \text{m} \]
\[ W = (0.16 \times 45.1 \, \text{N}) \times \cos 28° \times (2.8 \, \text{m}) \]
\[ W_f = 17.8 \, \text{J} \]](https://content.bartleby.com/qna-images/question/6c806aca-d4f4-488e-823b-e6a92c863c9f/2bd44cc4-5e5f-4b15-8bb6-debcc498ff98/eut4wm7_processed.png)
Transcribed Image Text:Below is a conservation of energy problem. The solution to this problem is provided. Assess whether the solution provided is correct or incorrect AND EXPLAIN WHY.
A 4.6 kg block is at rest against a horizontal spring that is compressed by 0.40 m. The spring has a spring constant of \( k_1 = 2750 \, \text{N/m} \). After leaving the spring, it travels up a 28° incline to a height of 2.8 m. At the top of the hill is a second spring with a spring constant of \( k_2 = 350 \, \text{N/m} \). The horizontal portions are frictionless, but the hill has a coefficient of kinetic friction equal to \( u_k = 0.16 \). The final velocity of the block is \( 9.78 \, \text{m/s} \).
**How much work is done by friction as the block slides up the hill?**
\[ W = F \cos \theta d \]
\[ F = F_k = u_k N \]
\[ u_k = 0.16 \]
\[ N = mg = (4.6 \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 45.1 \, \text{N} \]
\[ \theta = 28° \]
\[ d = 2.80 \, \text{m} \]
\[ W = (0.16 \times 45.1 \, \text{N}) \times \cos 28° \times (2.8 \, \text{m}) \]
\[ W_f = 17.8 \, \text{J} \]
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