Question

When being unloaded from a moving truck, a 16.0- kilogram suitcase is placed on a flat ramp inclined at 40.0 o. When released from rest, the suitcase accelerates down the ramp at 1.36 m/s². What is the coefficient of kinetic friction between the suitcase and the ramp?

Answer 1

To find the coefficient of kinetic friction between the suitcase and the ramp, you can use the equation: acceleration = g*sin(theta) - mu_k*cos(theta). Plugging in the given values, you can solve for mu_k.

Step-by-step explanation:

To find the coefficient of kinetic friction between the suitcase and the ramp, we can use the equation:

acceleration = g*sin(theta) - mu_k*cos(theta)

Where g is the acceleration due to gravity (9.8 m/s²), theta is the angle of the ramp (40.0°), and mu_k is the coefficient of kinetic friction. Plugging in the given values, we get:

1.36 m/s² = 9.8 m/s² * sin(40.0°) - mu_k * cos(40.0°)

Simplifying the equation, we find:

mu_k = (9.8 m/s² * sin(40.0°) - 1.36 m/s²) / cos(40.0°)

Calculating this gives us the coefficient of kinetic friction between the suitcase and the ramp.

Answer 2

0.65812

Step-by-step explanation:

m = Mass of suitcase = 16 kg

`\theta` = Incline angle = 40°

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block = 1.36 m/s²

f = Frictional force

The normal force is given by

`N=mgcos\theta`

In x direction

`mgsin\theta-f=ma\\\Rightarrow f=mgsin\theta-ma\\\Rightarrow f=16* 9.81* sin40-16* 1.36\\\Rightarrow f=79.13194\ N`

Frictional force is given by

`f=\mu N=\mu mgcos\theta\\\Rightarrow \mu=(f)/(mgcos\theta)\\\Rightarrow \mu=(79.13194)/(16* 9.81* cos40)\\\Rightarrow \mu=0.65812`

The coefficient of kinetic friction between the suitcase and the ramp is 0.65812