Internal Energy
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ENERGY PROBLEMS
When work is done on or by a system
Approximately how much kinetic energy
does the green box acquire?
PE = Just under 600 Nm
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Ex 1)
An object slides across a horizontal table which changes its KE from 20 J to
18 J in 2 sec.
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Ex 1) An object slides across a horizontal table which changes its kinetic energy from 20 J to 18 J in 2 sec. What work does the object do against friction?
Wf = 2 J
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**Work against friction increases Friction is a (can't get back) |
Joule Paddle
Conservation of Energy
Harper's New Monthly Magazine, No. 231, August, 1869.
Joule Paddle
Penn State Schuylkill
Work Energy Relationship
The work done on or by a system
= ΔKE + ΔPE + Wf
Wf - work done against friction
If ΔPE of yellow box = 30 J,
how much work did the green box do on it?
| Work done by the ball on the floor changes the ball's mechanical energy |
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When the WORK equation doesn't WORK, use an ENERGY equation. When an ENERGY equation doesn't WORK, use the WORK equation |
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Ex 2) A 10 kg frictionless cart is resting on a horizontal table. A force of 10 N is applied to the cart for a distance of 8 m. What is the carts new KE? |
Ex 2) A 10 kg frictionless cart is resting on a horizontal table. A force of 10 N is applied to the cart for a distance of 8 m. What is the carts new kinetic energy?
m = 10 kg
F = 10 N
d = 8 m
KE = (˝)mv2 ??
ΔKE = W
= Fd = 10N(8m)
= 80 Nm
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Ex 3) 100 Joules of work is needed to lift a 100 kg object from the ground. What is the objects new P.E.? |
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Ex 3) 100 Joules of work is needed to lift a 100 kg object from the ground.
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What is the objects new P.E. ? |
W = ΔPE = 100J
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Ex 4) Crate is pulled 6.00 m up an incline with a force of 50.0 N. P.E. of cart increases by 250. J. Total work done against friction? |
Ex 4) Crate is pulled 6.00 m up an incline with a force of 50.0 N. If the PE of the cart increases by 250. J. , what is the total work done against friction?
d = 6.00 m
F = 50.0 N
ΔPE = 250. J
W = Fd = 50.0 N(6 m)
= 300. J
W = ΔPE + Wf
300. J = 250. J + Wf
Wf = 50.0 J
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Ex 5) 4 N is exerted on a 1 kg mass at rest for 2 m, causing it to move
A. Find the objects K.E. |
KE = (1/2)mv2
Work on object = ΔKE
ΔKE = W = Fd = 4 N(2 m)
= 8 Nm
B. Find the objects velocity
KE = (1/2)mv2
8 Nm = (1/2)1 kg(v2)
16 = v2
V = 4 m/s
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Ex 6) A force of 100.0 N lifts the object 10.00 m off the ground. What is the object's new PE? What is the mass of the object? |
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Ex 6) A force of 100.0 N lifts the object 10.00 m off the ground. What is the object's new potential energy?
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mass of object? |
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ΔPE = W = Fd
= 100.0 N(10.00 m)
= 1000. Nm
What is the mass of the object?
PE = 1000. J = mgh
= m(10.00 m/s2)10.00 m = 1000. J
m = 10.00 kg
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Ex 7) Spring has a spring constant of 2.0 N/m. How much work must be done to stretch 5.0 m from its equilibrium position? |
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Ex 7) A spring has a spring constant of 2.0 N/m. How much work must be done to stretch it 5.0 m from its equilibrium position?
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k = 2.0 N/m x = 5.0 m W = ? |
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W = PE = 1/2kx2
= (1/2)2.0 N/m(5.0 m)2
= 25. J
Review
Name the energy conversion that
occurs in a Grandfather clock.
School blocks Youtube? Download file below.
St. Mary's Physics YouTube Channel
PE of the hanging weights
converts to the KE of the clock
Enrichment
Superfluids
The strange behavior of liquid helium
BBC
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