= 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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Chapter7: Work And Kinetic Energy
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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.
=
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.
How much work is done by friction as the block slides up the hill?
W = Fcosed
F = F₁ = U₁N
uk= 0.16
N = mg = (4.6 kg) * (9.8 m/s²):
0 = 28°
d = 2.80 m
= 45.1 N
W = (0.16 * 45.1 N) * cos28° * (2.8 m)
Wf = 17.8 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. = 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. How much work is done by friction as the block slides up the hill? W = Fcosed F = F₁ = U₁N uk= 0.16 N = mg = (4.6 kg) * (9.8 m/s²): 0 = 28° d = 2.80 m = 45.1 N W = (0.16 * 45.1 N) * cos28° * (2.8 m) Wf = 17.8 J
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