a medtech student pushes a 30kg carton of books across the floor of her dormitory by exerting a constant force of 185 N at an angle of 37 degrees with the horizon. If the opposing force of friction is 70 N, how much work does she do as she moves the carton 4 m in 15 s? What power in watts has she exerted?

To find the work done by the medtech student in moving the carton, we can use the formula for work, which is given by:

Work = Force x Distance x cos(angle)

Here, the force exerted by the student is 185 N, the distance covered is 4 m, and the angle with the horizon is 37 degrees. Plugging these values into the formula, we get:

Work = 185 N x 4 m x cos(37 degrees)

To calculate the value of cos(37 degrees), we can use a scientific calculator, or use the trigonometric identity cos(angle) = adjacent/hypotenuse. In this case, the adjacent side is the force acting parallel to the direction of motion (185 N x cos(37 degrees)), and the hypotenuse is the total force applied by the student (185 N). Substituting these values into the formula, we get:

cos(37 degrees) = (185 N x cos(37 degrees)) / 185 N

cos(37 degrees) = (adjacent side) / (hypotenuse)

cos(37 degrees) = (185 N x cos(37 degrees)) / 185 N

Simplifying, we find:

cos(37 degrees) = cos(37 degrees)

Therefore, the value of cos(37 degrees) is equal to 1.

Now, substituting this value back into the work formula, we have:

Work = 185 N x 4 m x 1

Work = 740 joules

So, the work done by the medtech student in moving the carton 4 m is 740 joules.

To calculate the power exerted by the student, we can use the formula for power, which is given by:

Power = Work / Time

Here, the work done is 740 joules, and the time taken is 15 s. Substituting these values into the formula, we get:

Power = 740 joules / 15 s

Using a calculator, we find:

Power ≈ 49.33 watts

Therefore, the power exerted by the medtech student is approximately 49.33 watts.

Is the angled force applied above or below horizontal? It makes a difference