Peoples Physics Concepts

http://www.ck12.org Chapter 8. Energy Conservation

Question: A pile driver lifts a 500 kg mass a vertical distance of 20 m in 1.1 sec. It uses 225 kW of supplied power
to do this.
a) How much work was done by the pile driver?
b) How much power was used in actually lifting the mass?
c) What is the efficiency of the machine? (This is the ratio of power used to power supplied.)
d) The mass is dropped on a pile and falls 20 m. If it loses 40,000 J on the way down to the ground due to air
resistance, what is its speed when it hits the pile?
Answer:
a) We will use the equation for work (which gives us the amount of energy transferred) and plug in the known
values to get the amount of work done by the pile driver.
W=F d=mgd=500kg× 9 .8m/s^2 ×20m= 9. 8 × 104 J

b) We will use the power equation and plug in the known values and then convert to kW at the end.

P=

W

∆t

=

9. 8 × 104 J

1 .1s

=89000W×

1kW

1000W

=89kW

c) Efficiency is defined as the Power out divided by the Power in. Thus, this is simply a division problem.

E f f=

power used

power supplied

=

89kW

225kW

=. 40

d) We have already solved for the amount of energy the mass has after the pile driver performs work on it (it has

9. 8 × 104 J). If on the way down it loses 40000J due to air resistance, then it effectively has

98000J−40000J=58000J

of energy. So we will set the kinetic energy equation equal to the total energy and solve for v. This will give us the

velocity of the mass when it hits the ground because right before the mass hits the ground, all of the potential

energy will have been converted into kinetic energy.

58000J=

1

2

mv^2 ⇒v=

√

58000J× 2

m

=

√

58000J× 2

500kg

= 15 .2m/s

Watch this Explanation

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/1902

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/354