Answer:
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
