Answer:
4.73 m/s²
Explanation:
The free body diagram for the bucket is
Then, the net force is equal to
Fnet = T - mg = ma
Where T is the tension of the rope, m is the mass, g is the gravity, and a is the acceleration. Solving for a, we get:
[tex]\begin{gathered} T-mg=ma \\ \\ a=\frac{T-mg}{m} \end{gathered}[/tex]Replacing T = 20.2 N, m = 1.39 kg, and g = 9.8 m/s², we get
[tex]\begin{gathered} a=\frac{20.2\text{ N - \lparen1.39 kg\rparen\lparen9.8 m/s}^2)}{1.39\text{ kg}} \\ \\ a=\frac{20.2\text{ N - 13.622 N}}{1.39\text{ kg}} \\ \\ a=\frac{6.578\text{ N}}{1.39\text{ kg}} \\ \\ a=4.73\text{ m/s}^2 \end{gathered}[/tex]Therefore, the acceleration is 4.73 m/s²
