How much work is it to push a box (mass 190 kg) up an incline (angle 16 degrees with the horizontal) that is 8.3 meters long, if the coefficient of kinetic friction between the box and the incline is 0.6?
To find the work done to push a box up an incline, we need to consider two main factors: the force required to overcome the gravitational force and the force required to overcome the frictional force.
Let's break down the problem into steps:
Step 1: Calculate the gravitational force acting on the box.
The force due to gravity can be calculated using the formula:
F_gravity = mass * acceleration due to gravity
where mass = 190 kg and acceleration due to gravity ≈ 9.8 m/s^2.
Plugging in these values:
F_gravity = 190 kg * 9.8 m/s^2
F_gravity = 1862 N
Step 2: Calculate the force required to lift the box along the incline.
The component of the gravitational force acting parallel to the incline can be found as:
F_parallel = F_gravity * sin(θ)
where θ is the angle of the incline (16 degrees).
Plugging in the values:
F_parallel = 1862 N * sin(16 degrees)
F_parallel ≈ 516.55 N
Step 3: Calculate the force of friction.
The force of friction can be found using the formula:
F_friction = coefficient of friction * normal force
The normal force is the force perpendicular to the incline, which is equal in magnitude to the component of the gravitational force acting perpendicular to the incline:
F_perpendicular = F_gravity * cos(θ)
Plugging in the values:
F_perpendicular = 1862 N * cos(16 degrees)
F_perpendicular ≈ 1767.25 N
Now, we can find the force of friction:
F_friction = 0.6 * 1767.25 N
F_friction ≈ 1060.35 N
Step 4: Calculate the net force required to move the box up the incline.
The net force required is the vector sum of the force required to lift the box and the force of friction acting in the opposite direction:
F_net = F_parallel + F_friction
F_net = 516.55 N + 1060.35 N
F_net ≈ 1576.9 N
Step 5: Calculate the work done.
The work done to push the box up the incline is given by the formula:
Work = force * distance * cosine(θ)
where distance = 8.3 meters and θ is the angle of the incline (16 degrees).
Plugging in the values:
Work = 1576.9 N * 8.3 m * cos(16 degrees)
Work ≈ 20130.55 Joules
Therefore, the work required to push the box up the incline is approximately 20130.55 Joules.
m g 8.3 sin 16 + friction force * 8.3
friction force
= .6 (m g cos 16)