A 225 kg box is loaded onto a truck by pulling it up an inclined plane (10 m hypotenuse difference) by a force that is parallel to the incline (T = 315 N). If the bed of the truck is 1.2 m above the ground what is the efficiency of this simple machine in loading the box into the truck?

lol

To calculate the efficiency of the inclined plane in loading the box into the truck, we need to determine the work input and work output.

Step 1: Find the work input.
The work input is the force exerted times the distance through which it is applied. In this case, the force exerted is equal to the pulling force, which is parallel to the incline (T = 315 N), and the distance is the length of the inclined plane (10 m).

Work input = Force × Distance
= T × d
= 315 N × 10 m
= 3150 N⋅m

Step 2: Find the work output.
The work output is the force applied to lift the box times the vertical distance it is lifted. We know the weight of the box (225 kg) and the height difference between the truck bed and the ground (1.2 m).

Work output = Force × Distance
= m × g × d
= 225 kg × 9.8 m/s² × 1.2 m
= 2646 N⋅m

Step 3: Calculate the efficiency.
The efficiency can be calculated as the work output divided by the work input, multiplied by 100 to convert it to a percentage.

Efficiency = (Work output / Work input) × 100
= (2646 N⋅m / 3150 N⋅m) × 100
= 0.84 × 100
= 84%

Therefore, the efficiency of the inclined plane in loading the box into the truck is 84%.

To find the efficiency of the simple machine in loading the box onto the truck, we first need to understand the concept of efficiency in relation to machines.

Efficiency is a measure of how effectively a machine converts input work to output work. It is expressed as a percentage and can be calculated using the following formula:

Efficiency = (Output Work / Input Work) x 100%

In this case, the output work refers to the work done by the inclined plane in lifting the box onto the truck, while the input work represents the work done by the person pulling the box.

To find the output work, we need to calculate the actual force exerted by the inclined plane to lift the box. This force can be determined by decomposing the applied force into two components: one parallel to the incline (the force that lifts the box) and another perpendicular to the incline (the force that pushes the box into the incline).

First, we can calculate the component of the applied force parallel to the incline:

Force_parallel = T × sin(θ)

Where T is the applied force (315 N) and θ is the angle of inclination of the plane. To find θ, we can use trigonometry:

sin(θ) = height difference / hypotenuse difference

The height difference is given as 1.2 m, and the hypotenuse difference is given as 10 m. Substituting these values into the formula, we get:

sin(θ) = 1.2 / 10

Next, we can calculate the force_parallel:

Force_parallel = 315 × (1.2 / 10)

Now, we can calculate the output work using the formula:

Output Work = Force_parallel × distance

The distance is the horizontal distance along the incline over which the box is lifted, which is equal to the hypotenuse difference of 10 m.

Output Work = Force_parallel × distance = (315 × (1.2 / 10)) × 10

Now, we need to calculate the input work. The input work is given by the product of the applied force and the distance over which it is applied:

Input Work = T × distance

Input Work = 315 × 10

Finally, we can calculate the efficiency using the formula mentioned earlier:

Efficiency = (Output Work / Input Work) × 100% = [(315 × (1.2 / 10)) × 10] / (315 × 10) × 100%

By simplifying the equation, we get:

Efficiency = (1.2 / 10) × 100%

Efficiency = 12%

Therefore, the efficiency of this simple machine in loading the box onto the truck is 12%.