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A 10 kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 5 N opposing the motion.
Calculate the acceleration of the object.

Respuesta :

indrawboy

Newton second's law

∑F = m.a

40 - 5 = 10.a

35 = 10.a

a = 3.5 m/s²

xKelvin

Answer:

The acceleration of the object was 3.5 m/s².

Explanation:

Let the direction of the horizontal force be positive.

By Newton's second law, the sum of the horizontal forces is as follows:
[tex]\displaystyle \sum F_x = F_A - F_k = ma[/tex]

Where Fk is the frictional force and FA is the horizontal force.

Substitute in known values and solve for acceleration a:

[tex]\displaystyle \begin{aligned} (40\text{ N}) - (5\text{ N}) & = (10\text{ kg})a \\ \\ a &= 3.5\text{ m/s$^2$}\end{aligned}[/tex]

Hence, the acceleration of the object was 3.5 m/s².

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