Answer:
a= 4.14 m/s²
Explanation:
We calculate the weight component parallel to the displacement of the block:
We define the x-axis in the direction of the inclined plane , 25° to the horizontal.
W= m*g : Total block weight
Wx= W*sen25°= m*g* sen25°
We apply Newton's second law :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Problem development
We apply the formula (1) to calculate the acceleration of the block:
∑Fx = m*a
Wx = m*a
m*g* sen25° = m*a : We divide by m on both sides of the equation
g* sen25° = a
a = g* sen25° = 9.8* sen25° = 4.14 m/s²
