Question

Use the equation for power to solve the following problems.

1. While rowing across the lake during a race, John does 3,960 J of work on the oars in 60.0s. What is his power output in watts?

2. Anna walks up the stairs on her way to class. She weighs 565N, and the stairs go up 3.25 m vertically.

a. If Anna climbs the stairs in 12.6 S, what is her power output?

b. What is her power output if she climbs the stairs in 10.5 s?

Answer 1

1) W=Pt
3,960=P x 60
3960/60=P
P=66
66W

2a) W=Fs
W=565x3.25
W=1836.25J
W=Pt
1836.25=P x 12.6
1836.25/12.6=P
Power output= 145.73W

2b)
1836.25/10.5=174.88W

Answer 2

The equation for power is P = W/t, where P is power in watts, W is work in joules, and t is time in seconds. Plugging in the values given in the problem, we get:

P = W/t = 3,960 J / 60.0 s = 66.0 W

Therefore, John's power output is 66.0 watts.

The equation for power is P = W/t, where P is power in watts, W is work in joules, and t is time in seconds. To solve for power, we need to find the work done and the time it takes to do that work.

a. The work Anna does when climbing the stairs is equal to the force she exerts (her weight) times the distance she travels vertically. Using the equation W = Fd, where W is work, F is force, and d is distance, we get:

W = Fd = (565 N)(3.25 m) = 1836.25 J

The time it takes Anna to climb the stairs is given as 12.6 s. Plugging in the values, we get:

P = W/t = 1836.25 J / 12.6 s = 145.48 W

Therefore, Anna's power output when climbing the stairs in 12.6 s is 145.48 watts.

b. If Anna climbs the stairs in 10.5 s, the time is different but the work done is the same, so we can use the same value for W:

W = 1836.25 J

Plugging in the new time, we get:

P = W/t = 1836.25 J / 10.5 s = 174.88 W

Therefore, Anna's power output when climbing the stairs in 10.5 s is 174.88 watts.