Question

1. a 12.7 kg video cart is rolled down a hallway. it is accelerated at 0.6 m/s^2. if a force of 18 n is applied to move the cart, what force of friction do the wheels have against the floor?

Explanation:

Step1: Apply Newton's second law

Newton's second law is \(F_{net}=ma\), where \(F_{net}\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. The net force \(F_{net}\) is the applied force \(F_{applied}\) minus the frictional force \(F_f\), so \(F_{net}=F_{applied}-F_f\).

Step2: Substitute the values

We know \(m = 12.7\ \text{kg}\), \(a=0.6\ \text{m/s}^2\), and \(F_{applied}=18\ \text{N}\). Substituting into \(F_{net}=ma\) gives \(F_{applied}-F_f=ma\). Then \(F_f=F_{applied}-ma\).

Step3: Calculate the frictional force

Substitute the values into the formula \(F_f = 18-(12.7\times0.6)\). First, calculate \(12.7\times0.6 = 7.62\). Then \(F_f=18 - 7.62=10.38\ \text{N}\).

Answer:

\(10.38\ \text{N}\)