Question

question 3. the diagram shows a 4.0 kg object accelerating at a rate of 10.0 m/s2 on a rough horizontal surface. based on the diagram, what is the magnitude of the frictional force ff in newtons acting on the object?

Explanation:

Step1: Apply Newton's second law

According to Newton's second law \(F_{net}=ma\), where \(F_{net}\) is the net force, \(m\) is the mass of the object, and \(a\) is the acceleration. The net force \(F_{net}=F_{applied}-F_{f}\) (assuming the applied force \(F_{applied}\) is in the direction of motion and frictional force \(F_{f}\) is opposite to the direction of motion)

So, \(F_{applied}-F_{f}=ma\)

Step2: Rearrange the formula to solve for \(F_{f}\)

We can rewrite the formula from Step 1 as \(F_{f}=F_{applied}-ma\)

Given \(m = 4.0\space kg\), \(a=10.0\space m/s^{2}\), and \(F_{applied}=50\space N\)

Substitute the values into the formula: \(F_{f}=50 - 4\times10\)

Step3: Calculate the value of \(F_{f}\)

First, calculate \(4\times10 = 40\)

Then, \(F_{f}=50 - 40\)

Answer:

\(10\space N\)