Question

A machine at a post office sends packages out a chute and down a ramp to be loaded into delivery vehicles. Calculate the acceleration of a box (in m/s²) heading down a 12.5° slope, assuming the coefficient of friction for a parcel on waxed wood is 0.100. (Enter the magnitude.)

A) 9.8 m/s²
B) 8.5 m/s²
C) 7.2 m/s²
D) 6.4 m/s²

Answer 1

The acceleration of the box heading down a 12.5° slope is approximately 8.5 m/s².

Step-by-step explanation:

To calculate the acceleration of the box heading down a 12.5° slope, we need to consider the forces acting on the box. The force of gravity can be broken down into two components: the component parallel to the slope (mg*sin(theta)), and the component perpendicular to the slope (mg*cos(theta)). The frictional force opposing the motion of the box is given by the coefficient of friction (mu) multiplied by the normal force (mg*cos(theta)). The net force acting on the box is the difference between the parallel component of gravity and the frictional force.

Using Newton's second law of motion (F = ma), we can set up an equation: net force = ma. Solving for acceleration, we get: a = (mg*sin(theta) - mu*mg*cos(theta))/m. Plugging in the values, the acceleration comes out to be approximately 8.5 m/s².