A 5.5 kg object is released at the top of an inclined plane that makes an angle

Question

A 5.5 kg object is released at the top of an inclined plane that makes an angle of 35 degrees above the horizontal. What is the acceleration if the coefficient of kinetic friction is 0.24?
A. 2.2 m/s^2 B. 1.0 C. 5.8 D. 7.4 E. 3.7
All are m/s^2 A 5.5 kg object is released at the top of an inclined plane that makes an angle of 35 degrees above the horizontal. What is the acceleration if the coefficient of kinetic friction is 0.24?
A. 2.2 m/s^2 B. 1.0 C. 5.8 D. 7.4 E. 3.7
All are m/s^2
A. 2.2 m/s^2 B. 1.0 C. 5.8 D. 7.4 E. 3.7
All are m/s^2

Explanation / Answer

Normal N = mgcos(th)

So, frictional force f = uN = umgcos(th)

Summing forces along the slope and applying Newton's 2nd law of motion:

F - f = ma

=> mgsin(th) - umgcos(th) = ma

=> gsin(th) - ugcos(th) = a (m cancelled from both sides)

=> a = 9.81 * (sin35 - 0.24*cos35)

= 3.7 m/s^2

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