A 5-kg block slides down a plane inclined at 30 0 to the horizontal. Find a. The acceleration of the block if the plane is frictionless.
b. The acceleration if the coefficient of kinetic friction is 3 2 .

a) 4.9m/s²

b) 2.18m/s²

Explanation:

a) According to Newton's second law of motion

\sum Fx = ma

Fm-Ff = ma

Fm is the moving force = Wsin theta

Ff is the frictional force = 0N (frictionless plane)

m is the mass

a is the acceleration

Substituting into the formula

Fm -Ff = ma.

Wsintheta = ma

mgsintheta = ma

gsintheta = a

a = 9.8sin30°

a = 9.8(0.5)

a = 4.9m/s²

Hence the acceleration of the block if the plane is frictionless is 4.9m/s²

b) Let the coefficient if friction given be 0.32

From the formula

Fm-Ff = ma

mgsintheta - nmgcostheta = ma

gsintheta - ngcostheta = a

9.8(sin30)-0.32(9.8)cos30 = a

4.9-2.72 = a

a = 2.18m/s²

Hence the acceleration if the coefficient of kinetic friction is 0.32 is 2.18m/s²

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