Question

Calculate the average power required to lift a 490-newton object a vertical distance of 2.0 meters in 10. seconds. [Show all work, including the equation and substitution with units.]

Answer 1

The average power required to lift a 490-newton object a vertical distance of 2.0 meters in 10 seconds is 98 Watts, found by dividing the work (980 Joules) by the time (10 seconds).

Step-by-step explanation:

To calculate the average power required to lift a 490-newton object a vertical distance of 2.0 meters in 10 seconds, you can use the formula:

Power (P) = Work (W) / Time (t)

Since Work (W) = Force (F) × Distance (d), and the force here is the weight of the object, you can substitute in the values:

Force (F) = 490 N

Distance (d) = 2.0 m

Time (t) = 10.0 s

So the work done is:

W = 490 N × 2.0 m = 980 J

Then divide this work by the time to find the power:

P = 980 J / 10 s = 98 W

The average power required to lift the object is 98 Watts.

Answer 2

For this problem, you should be able to differentiate the variables presented from each other in order to substitute them in their corresponding places in the formula or formulas to be utilized in this problem. As for this problem, the only formula to be utilized would be the formula for power which is force multiplied to distance over time or simply have force multiplied to speed since speed is equal to distance over time.

The formula would like this:

Power = force x distance / time Power = force x speed

P = 490 N x 2 m / 10 s P = 490 N x (2 m / 10 s)
P = 980 N m / 10 s P = 490 N x 0.2 m / s
P = 98 W P = 98 W

So the average power required to lift a 490-newton object a vertical distance of 2.0 meters in 10 seconds would be 98 watts.