Power question
A 60kg woman runs up a staircase 15m high in 20s.
a) How much power does she expend?
b) What is her horsepower rating?
a) How much power does she expend?
b) What is her horsepower rating?
Solution:
a) P = W/Δt The time rate of doing work. The amount of work divided by change in time. In order to find
w = mgh power, we need to know the number of watts it took and the amount of time it took. Watts
P = mgh/ Δt equals mass times gravity times height. We were given: m = 60 kg
= (60kg)(9.8 m/s)(15m)/20s g = 9.8 m/s
h = 15m
Δt = 20s
b) 1 hp = 746 W
440/746 = .59hp
We know that 1 hp equals 746w, we divide 440 by 746.
a) P = W/Δt The time rate of doing work. The amount of work divided by change in time. In order to find
w = mgh power, we need to know the number of watts it took and the amount of time it took. Watts
P = mgh/ Δt equals mass times gravity times height. We were given: m = 60 kg
= (60kg)(9.8 m/s)(15m)/20s g = 9.8 m/s
h = 15m
Δt = 20s
b) 1 hp = 746 W
440/746 = .59hp
We know that 1 hp equals 746w, we divide 440 by 746.
Conservation of Energy
A .5kg ball thrown upward has an initial K of 80J.
a) What are it's kinetic and potential energies when it has traveled 3/4 of the distance to its maximum height?
b) What is the ball's speed at this point?
c) What is its potential energy at its maximum height?
Solution:
a) U at 3/4 height = 3/4 mgh At the top of the maximum height, it's potential energy is 80J. At 3/4's of the way, it would be
3/4(80) = 60J 3/4(80).
Since 60J out of the 80J is transferred to potential energy, that means the other 20J is for kinetic energy because of it's motion in the air.
b) K = .5mv²
v = sqrt(2k/m)
V at 3/4 height = sqrt(2(20)/.5) = 8.9 m/s
We need to re-write the equation for kinetic energy, .5mv², solving for v. Since we are asked for the velocity at 3/4's of the way, we use the kinetic energy at that point.
c) Since there is no motion at the max point, there will only be potential energy so it equals 80J.
a) What are it's kinetic and potential energies when it has traveled 3/4 of the distance to its maximum height?
b) What is the ball's speed at this point?
c) What is its potential energy at its maximum height?
Solution:
a) U at 3/4 height = 3/4 mgh At the top of the maximum height, it's potential energy is 80J. At 3/4's of the way, it would be
3/4(80) = 60J 3/4(80).
Since 60J out of the 80J is transferred to potential energy, that means the other 20J is for kinetic energy because of it's motion in the air.
b) K = .5mv²
v = sqrt(2k/m)
V at 3/4 height = sqrt(2(20)/.5) = 8.9 m/s
We need to re-write the equation for kinetic energy, .5mv², solving for v. Since we are asked for the velocity at 3/4's of the way, we use the kinetic energy at that point.
c) Since there is no motion at the max point, there will only be potential energy so it equals 80J.
Work
A particular spring has a force constant of 2500 N/m.
a) How much work is done in stretching the related string by 6.0 cm.
b) How much more work is done in stretching the spring an additional 2.0 cm.
Solution:
a) .5kx² Since work in this case equals the sprint potential energy, we solve for it with the given values.
.5(2500 N/m)(.06m)² The work done by the applied force in compressing a string equals half of the spring constant times
= 4.5J the amount of stretch squared.
b) .5(2500 N/m)(.8m)² - .5(2500 N/m)(.06m)²
8 - 4.5 = 3.5 J
Pulling the spring 2 more cm makes it go 8cm away from starting point. To find the work done, we subtract the amount of work done by pulling it 6cm from work done by pulling it 8cm.
a) How much work is done in stretching the related string by 6.0 cm.
b) How much more work is done in stretching the spring an additional 2.0 cm.
Solution:
a) .5kx² Since work in this case equals the sprint potential energy, we solve for it with the given values.
.5(2500 N/m)(.06m)² The work done by the applied force in compressing a string equals half of the spring constant times
= 4.5J the amount of stretch squared.
b) .5(2500 N/m)(.8m)² - .5(2500 N/m)(.06m)²
8 - 4.5 = 3.5 J
Pulling the spring 2 more cm makes it go 8cm away from starting point. To find the work done, we subtract the amount of work done by pulling it 6cm from work done by pulling it 8cm.