How much work is it to push a box (mass 120 kg) up an incline (angle 16 degrees with the horizontal) that is 9.6 meters long, if the coefficient of kinetic friction between the box and the incline is 0.5?
To find out how much work it takes to push a box up an incline, we need to calculate the net work done on the box. The net work is equal to the product of force applied and the displacement of the box.
First, we need to find the gravitational force acting on the box. The gravitational force can be calculated using the formula:
force_gravity = mass * gravity
Given that the mass of the box is 120 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the gravitational force:
force_gravity = 120 kg * 9.8 m/s^2 = 1176 N
Next, we need to determine the force required to overcome friction. The force of friction can be calculated using the formula:
force_friction = coefficient_of_friction * force_normal
The force_normal is the component of the gravitational force acting perpendicular to the incline. It can be calculated using the formula:
force_normal = force_gravity * cos(angle)
Since the angle of the incline is given as 16 degrees, we can calculate the force_normal:
force_normal = 1176 N * cos(16 degrees) = 1130.7 N
Now, we can calculate the force of friction:
force_friction = 0.5 * 1130.7 N = 565.4 N
To push the box up the incline, we need to overcome both the force of gravity and the force of friction. The net force required to move the box can be calculated using the formula:
net_force = force_gravity + force_friction
net_force = 1176 N + 565.4 N = 1741.4 N
Finally, we can calculate the work done by multiplying the net force by the displacement along the incline:
work = net_force * displacement
The displacement is given as 9.6 meters. Substituting the values, we get:
work = 1741.4 N * 9.6 m = 16734.24 J
Therefore, it takes approximately 16734.24 Joules of work to push the box up the incline.