À Block slides down on an inclined plane of 30degree with an acceleration equal to g/4 .find the coefficient of kinetic energy
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Answered by
112
Given :Inclination angle=θ=30 degreesa=g/4 m/s²Coefficient of Kinetic friction .
Total net force= mg/4=mgsinθ-μmgcosθ
where μ is coefficient of kinetic friction.
mg/4=mg[sinθ-μcosθ]
[sinθ-μcosθ] =1/4
Substituting the inclination angle, we get
sin30- μcos30=1/4
1/2-μ√3/2=1/4
1/2-1/4=μ√3/2
2-1/4=μ√3/2
1/4=μ√3/2
μ=1/√3/2
Total net force= mg/4=mgsinθ-μmgcosθ
where μ is coefficient of kinetic friction.
mg/4=mg[sinθ-μcosθ]
[sinθ-μcosθ] =1/4
Substituting the inclination angle, we get
sin30- μcos30=1/4
1/2-μ√3/2=1/4
1/2-1/4=μ√3/2
2-1/4=μ√3/2
1/4=μ√3/2
μ=1/√3/2
Answered by
49
Let the cofficent of kinetic friction be u
( Force Total ) mg / 4 = mg sin Ф - u m g cos Ф
Solve to get the answer as follows:
sin - u cos = 1 / 4
Substitute : Ф = 30 we get
u = 2 / √3
Please mark the answer as brainliest thank you
( Force Total ) mg / 4 = mg sin Ф - u m g cos Ф
Solve to get the answer as follows:
sin - u cos = 1 / 4
Substitute : Ф = 30 we get
u = 2 / √3
Please mark the answer as brainliest thank you
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