This question is incomplete
Complete Question
m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.
a) if the system is released from rest what will be its acceleration
Answer:
0.7 m/s²
Explanation:
The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.
(a) if the system is released from rest what will be its acceleration
g = acceleration due to gravity = 9.81 m/s²
Coefficient of Kinetic Friction = μk = 0.30
m1 = 10kg
m2 = 4.0kg
The formula to solve question a is given as:
a = acceleration at rest
m2g- μk m1g = (m1+ m2) a
Making a the subject of the formula:
a = (m2g- μk×m1g )/(m1+ m2)
a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)
a = 0.7 m/s²
