A med tech student pushes a 30-kg 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 horizontal. If the opposing force of friction is 70 N, how much work does he do as he moves the carton in 4m in 15s? What power in watts has he expended ?
the final velocity of the carton is not given, so assume zero.
forceexertedhorizontal=185cos37
distance=4m
workexerted=forcehorizontal*distance
power=work/timeinseconds
Well, well, well, it seems like our med tech student has gotten some serious workout! Let's calculate the work done and the power expended, shall we?
First, let's break down the force into horizontal and vertical components. The horizontal component of the force can be calculated using cosine:
Horizontal force = 185 N * cos(37°)
Now, we can calculate the work done by multiplying the horizontal force by the distance:
Work done = Horizontal force * distance
Power is simply the rate at which work is done, so we can find it by dividing the work done by the time taken:
Power = Work done / time
So, let's crunch the numbers!
Horizontal force = 185 N * cos(37°)
Work done = Horizontal force * 4 m
Power = Work done / 15 s
Now, if you'll excuse me, I'll go run some calculations. *beep boop beep*
After some number-crunching:
Horizontal force = 185 N * cos(37°) ≈ 148.592 N
Work done = Horizontal force * 4 m ≈ 594.368 J
Power = Work done / 15 s ≈ 39.625 W
So, our muscly med tech student has done approximately 594.368 J of work and expended about 39.625 W of power. Quite an impressive feat, don't you think?
To calculate the work done by the student as she moves the carton, we can use the formula:
Work = force * displacement * cos(theta)
where:
force = the magnitude of the applied force = 185 N
displacement = the distance moved = 4 m
theta = the angle between the force and displacement vectors = 37 degrees
Let's calculate the work done:
Work = 185 N * 4 m * cos(37 degrees)
Work = 740 N*m * cos(37 degrees)
Work = 740 N*m * 0.7986
Work ≈ 590.32 Joules
Therefore, the work done by the student to move the carton is approximately 590.32 Joules.
To calculate the power expended by the student, we can use the formula:
Power = work / time
where:
work = the work done = 590.32 Joules
time = the time taken = 15 s
Let's calculate the power expended:
Power = 590.32 Joules / 15 s
Power ≈ 39.35 Watts
Therefore, the power expended by the student is approximately 39.35 Watts.
To find the work done by the med tech student, we can use the formula: work = force x distance x cos(theta), where theta is the angle between the force and the direction of displacement.
First, let's calculate the horizontal component of the applied force:
horizontal force = force x cos(theta)
horizontal force = 185 N x cos(37 degrees)
horizontal force ≈ 148.47 N
Since the applied force and the direction of displacement are in the same direction (horizontal), the angle theta is 0 degrees. Therefore, we can simplify the work formula to:
work = force x distance
work = 148.47 N x 4 m (since the distance is in the horizontal direction)
work ≈ 593.88 Joules
The work done by the med tech student to move the carton 4 meters is approximately 593.88 Joules.
To find the power expended by the med tech student, we can use the formula: power = work / time
power = 593.88 J / 15 s
power ≈ 39.59 Watts
The power expended by the med tech student is approximately 39.59 Watts.