Question

if the two forces shown in the diagram below are applied to the 2 kg block, then what is the resulting acceleration?
f₁ = 2.0 n
2.0 kg
f₂ = 8.0 n
frictionless surface
a 5.0 m/s² to the right
b 5.0 m/s² to the left
c 3.0 m/s² to the left
d 3.0 m/s² to the right

Explanation:

Step1: Calculate the net force

The forces are in opposite directions. Let the right - direction be positive. The net force \(F_{net}=F_2 - F_1\). Given \(F_1 = 2.0N\) (left, so negative in our sign - convention) and \(F_2=8.0N\) (right, positive). Then \(F_{net}=8.0N-2.0N = 6.0N\) (positive, so to the right).

Step2: Use Newton's second law \(F = ma\) to find acceleration

Newton's second law is \(a=\frac{F_{net}}{m}\). We know \(m = 2.0kg\) and \(F_{net}=6.0N\). Substitute the values: \(a=\frac{6.0N}{2.0kg}\).

Answer:

\(d\). \(3.0m/s^{2}\) to the right