A woman pushes a crate along the floor of a warehouse with a force of 31.9 N directed downward at an angle of 23.6° from the horizontal. If the crate has a mass of 11.9 kg and the coefficient of friction is 0.0975, find...

(a) the force of friction
b. (b) theacceleration

Answer sent to repost later

http://www.jiskha.com/display.cgi?id=1326928884

To find the force of friction and the acceleration, we can use Newton's second law and the equation for friction.

Let's break down the problem into components. The force of 31.9 N can be split into its horizontal and vertical components.

Vertical Component:
The vertical component of force is given by:
F_vertical = F * sin(theta)
F_vertical = 31.9 N * sin(23.6°)
F_vertical = 12.85 N

The vertical component does not affect the horizontal motion of the crate because the floor provides an equal and opposite normal force, canceling out the vertical component.

Now, let's focus on the horizontal component:

Horizontal Component:
The horizontal component of force is given by:
F_horizontal = F * cos(theta)
F_horizontal = 31.9 N * cos(23.6°)
F_horizontal = 28.84 N

The force of friction can be calculated using the equation:
Friction force = coefficient of friction * normal force

To find the normal force:
Normal force = mass * gravitational acceleration
Normal force = 11.9 kg * 9.8 m/s²
Normal force = 116.62 N

Now we can calculate the force of friction:
Friction force = 0.0975 * 116.62 N
Friction force = 11.37 N (approximately)

To find the acceleration, we can use Newton's second law:
Force_net = mass * acceleration

The net force in the horizontal direction is the difference between the applied force and the force of friction:
Force_net = F_horizontal - Friction force
Force_net = 28.84 N - 11.37 N
Force_net = 17.47 N

Now we can find the acceleration:
17.47 N = 11.9 kg * acceleration
acceleration = 17.47 N / 11.9 kg
acceleration = 1.47 m/s² (approximately)

So, the answers are:
(a) The force of friction is approximately 11.37 N.
(b) The acceleration is approximately 1.47 m/s².