# physics

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.45 and μk = 0.38, respectively.
The acceleration of gravity is 9.8 m/s2 .
What is the frictional force acting on the 25 kg mass?
What is the largest angle which the incline can have so that the mass does not slide down the incline?
What is the acceleration of the block down the incline if the angle of the incline is 29◦ ?

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1. a. Wt. = m * g 25kg * 9.8N/kg = 245 N.

F1 = Force parallel to the incline.
F2 = Force perpendicular to the incline.

Q1. F1 = 245*sinA = u*F2
245*sinA = 0.45*245*cosA
Divide both sides by 245*cosA:
sinA/cosA = 0.45
Replace SinA/cosA with tanA:
tanA = 0.45
A = 24.2o = Angle of the incline.

Fs=u*245*cosA=0.45*245*cos24.2=100.6 N.
= Force of static friction.

Q2. The largest angle is the angle at which the force of static friction is
equal to F1. The angle is 24.2o.

Q3. F1 = 245*sin29 = 118.8 N.
F2 = 245*cos29 = 214.3 N.

Fk = u*F2 = 0.38 * 214.3 = 81.4 N. =
Force of kinetic friction.

a=Fn/m = (F1-Fk)/m=(118.8-81.4)/25 = 1.49 m/s^2.

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2. this is sd afae[a

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3. 24.2o what is the sign after 24.2?

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4. degrees probably

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