Answer:
aceleration as start the move is 2.45 m/s²
Step-by-step explanation:
We use Newton's second law on each axis. In the x-axis we have two forces the driving force and the friction force in the opposite direction, in the axis and we also have two forces the normal one up and the weight down.
To find the driving force we use the equilibrium condition just before the movement begins, in this case the friction coefficient is static.
X-axis F-fr = 0 ⇒ F = fr
Y-axis N-w = 0 ⇒ n = w = mg
The expression for the friction force is fr = μ N
Fr = μs m g
F = μs m g
μs is the coefficient of static friction
Now it starts moving and some of the links are broken, which is reflected in the fact that the coefficient of kinetic friction is lower, if we keep in value of the driving force, Newton's second law remains
F -fr = ma
N-w = 0 ⇒ N = w = mg
fr = μk N = μk m g
μk is the coefficient of kinetic friction
a = (F -fr) / m
a = (μs m g - μk m g) / m
a = (μs - μk) g
a = (0.4 - 0.15) 9.8
a = 2.45 m/s²
This is the acceleration due to the decrease in the friction coefficient