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Answer: A block with mass 0.50 kg is forced against a
Chapter 7, Problem 43P(choose chapter or problem)
A block with mass 0.50 \(\mathrm{kg}\) is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m (Fig. P7.43). When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant is 100 N/m. What is the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the block and the tabletop?
Equation Transcription:
Text Transcription:
kg
Mu k
k=100 N/m
m=0.50 kg
0.20 m
1.00 m
Questions & Answers
QUESTION:
A block with mass 0.50 \(\mathrm{kg}\) is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m (Fig. P7.43). When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant is 100 N/m. What is the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the block and the tabletop?
Equation Transcription:
Text Transcription:
kg
Mu k
k=100 N/m
m=0.50 kg
0.20 m
1.00 m
ANSWER:
Solution 43P
Step 1:
Elongation of the spring, x = 0.20 m
Force constant of the spring, k = 100 N/m
The potential energy of the spring, PE = ½ kx2 = ½ ×100 N/m × 0.202 m2 = 2 J
The equation for frictional force, F = 𝛍k mg
Where, 𝛍k - coefficient of static friction
m - Mass of the body
g - Acceleration due to gravity
The work done by the spring, W = FX
Where,X - Distance of travel
Mass, m = 0.5 kg
Distance, X = 1 m
Therefore, W = 𝛍k mg X = 𝛍k × 0.5 kg × 9.8 m/s2 × 1 m = 4.9 𝛍k J