= 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 k₁ 2750 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 k₂ = 350 N/m. The horizontal portions are frictionless, but the hill has a coefficient of kinetic friction equal to uk= 0.16. The final velocity of the block is 9.78 m/s.

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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?**

\[ W = F \cos \theta d \]

\[ F = F_k = u_k N \]

\[ u_k = 0.16 \]

\[ N = mg = (4.6 \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 45.1 \, \text{N} \]

\[ \theta = 28° \]

\[ d = 2.80 \, \text{m} \]

\[ W = (0.16 \times 45.1 \, \text{N}) \times \cos 28° \times (2.8 \, \text{m}) \]

\[ W_f = 17.8 \, \text{J} \]
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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