Question

1. calculate the force of friction acting on an 1800 kg car if the driver brings the car from 100 km/h to rest over a distance of 40.0 m.
2. a 2.3 kg bag of candy is pulled to the west with a force of 12.0 n and to 30.0° south of east with a force of 15.0 n. calculate the net force acting on the bag of candy.
3. the coefficient of static friction between a 10.0 kg block and the floor is 0.60, while the coefficient of kinetic friction between the same two materials is 0.40. determine the force of friction acting between the two materials when the following forces are applied:

a. 20 n
b. 50 n
c. 80 n
d. 100 n

Explanation:

Step1: Convert initial velocity units

Convert \(v_i = 100\ km/h\) to \(m/s\). Using the conversion factor \(1\ km = 1000\ m\) and \(1\ h=3600\ s\), we have \(v_i = 100\times\frac{1000}{3600}\approx27.78\ m/s\), \(v_f = 0\ m/s\), and \(d = 40.0\ m\).

Step2: Use the kinematic equation \(v_f^2=v_i^2 + 2ad\) to find acceleration \(a\)

Rearrange the equation for \(a\): \(a=\frac{v_f^2 - v_i^2}{2d}\). Substitute the values: \(a=\frac{0-(27.78)^2}{2\times40.0}\).

$$a=\frac{-771.7284}{80}\approx - 9.65\ m/s^2$$

Step3: Use Newton's second law \(F = ma\) to find the frictional force

Given \(m = 1800\ kg\) and \(a=-9.65\ m/s^2\), then \(F_f=ma\).

$$F_f=1800\times(- 9.65)=-17370\ N$$

The negative sign indicates the force is in the opposite direction of motion.

Answer:

The force of friction is \(17370\ N\) (opposite to the direction of the car's motion).