Work and

Power

Work: (W) the transfer of energy that occurs

when a force pushes on a moving object.•Object must move

• Force must be applied in the direction of the movement.

Measured in Joules

Work can be calculated by:W = work

Measured in J

W = FdF = force

Measured in N

d = distanceMeasured in m

You move a 75 kg refrigerator 35 m. This requires a force of 90 N.

How much work, in joules, was done while moving the refrigerator?

m = 75 kg

d = 35 m

F = 90 N

W = ?

W = FdW = (90)(35)

W = 3,150 J

The Potential Energy of an object is equal to the amount of work that it can do.

Gravitational Potential EnergyGPE = m*g*h ( mass * gravity * height )since f = ma (mass * acceleration)(mass * 9.8) = force height = distance

GPE = mgh = fdWork = fd

The Kinetic Energy of an object is equal to the amount of work that it can do.

KE = ½ mv2

If you know the KE then you know the amount of work that must be done on the object to stop it.

KE = Work = fd

1.A book weighing 1 N is lifted 2 m. How much work was done?

W = 2 J

W = Fd

W = 1*2

A 1 kg book is lifted 2 m.

How much work was done?

W = 19.6 J

W = Fd

W = 9.8 * 2

F = m * a 9.8 = 1 * 9.8

250 J of work are needed to push a box up a ramp. The force needed to push the box up the ramp is 10 N.

How far up the ramp was the box pushed?

d = 25 m

W = Fd

250 = 10 * d

If 56 J of work energy is required to lift a dumbbell to a height of 7 m.

What is the weight of the dumbbell?

F = 8 N

W = Fd

56 = F * 7

P = power

Measured in J/s

Or

Watts = W

𝑷 =𝑾

𝒕

W = work

Measured in J

t = time

Measured in s

Power: (P) The rate at which work is done.

It took 150 seconds to move a refrigerator.

You did 3,150 joules of work in the process.

How much power was required to move the refrigerator?

t = 150 s

W = 3,150 J

P = ? P = 3150

150

P = 21 J/s (21 Watts)

𝑷 =𝑾

𝒕

work and power Bozeman

work and power Dave

work and power crash course

GPE = m*g*h = work = f*dHow much work must be done on a 20kg object to raise it 10 meters

How much GPE will it have?GPE = m*g*h GPE = 20 * 9.8 * 10 = 1,960 J

GPE = Work = (1,960 J)

How fast will a 20kg object be going when it hits the ground if it is dropped from 10 meters highAssuming there is no loss due to friction, all the energy is converted into KE

GPE = KEGPE = m*g*h = 20 * 9.8 * 10 = 1,960 Jsince all the energy was converted KE also equals 1960 JKE = ½ mv2 1960 = ½ mv2 1960 = ½ *20*v2

2 ∗1960

20= 14 m/s

A 3000 kg car going 120 km/h slams on its breaks and skids to a stop after sliding 44 meters.

What was the force exerted on the car during the skid?

KE = ½ mv2 = ½ * 3000 * ? = 1,663,335 J = ( 1.66 x 107 )

120 𝑘𝑚

1ℎ𝑟∗1000m

1km∗

1hr

3600s= 33.3

m

s

𝟑𝟑. 𝟑𝟐

The amount of work done on the car is equal to the amount of lost KE. 1.66 x 107 J = Work = fd

1.66 x 107 J = f * 44 f = 37,803 N

A 3000 kg car going 120 km/h slams on its breaks and skids to a stop after sliding 44 meters.

What was the force exerted on the car during the skid?

f = 37,803 N

𝟑𝟑. 𝟑𝟐 m/s

𝑣2 = 2𝑎𝑑 33.32 = 2 ∗ 𝑎 ∗ 44 12.6 𝑚/𝑠2 = 𝑎

𝑓 = 𝑚𝑎 𝑓 = 3000 ∗ 12.6𝑎

We can also find the answer using basic kinematic equations

WorksheetRS pg. 28