Question

1. the coefficient of kinetic friction acting between the bottom of a 10.0 kg box and the floor is 0.30. if the box is pulled to the right by a force of 75.0 n, determine the acceleration of the box.
2. calculate the acceleration of the box below. assume a 25.0 n force of friction on box b.
3. calculate the work done on a 1500 kg car in accelerating it from rest to 30.0 m/s. if the force that is used to accelerate the car is 3000 n, over what distance was the work done?

Explanation:

Step1: Analyze forces

According to Newton's second law \(F_{net}=ma\). The net force \(F_{net}\) is the applied force \(F\) minus the frictional force \(F_f\). The frictional force \(F_f=\mu_k N\), and since \(N = mg\) (normal force equals weight when no vertical acceleration), \(F_f=\mu_kmg\).

Step2: Calculate net force

Given \(m = 10.0\space kg\), \(\mu_k=0.30\), \(g = 9.8\space m/s^2\), \(F = 75.0\space N\). First, calculate \(F_f\):
\(F_f=\mu_kmg=0.30\times10.0\times9.8 = 29.4\space N\)
Then, \(F_{net}=F - F_f=75.0-29.4 = 45.6\space N\)

Step3: Calculate acceleration

Using \(F_{net}=ma\), we can solve for \(a\). Rearranging gives \(a=\frac{F_{net}}{m}\)
Substituting \(F_{net}=45.6\space N\) and \(m = 10.0\space kg\)
\(a=\frac{45.6}{10.0}=4.56\space m/s^2\)

Answer:

\(4.56\space m/s^2\)