ENERGY PROBLEMS
When work is done on or by a system
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?
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 |
Work Energy Relationship
The work done on or by a system
= ΔKE + ΔPE + Wf
Wf - work done against friction
Joule Paddle - Penn State Schuylkill
| When the WORK equation doesn't WORK, use an ENERGY equation. When an ENERGY equation doesn't WORK use the WORK equation |
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?
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 = (½)mv2 ???
ΔKE =
W = Fd = 10N(8m)
= 80 Nm
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. ? |
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 = 100 J
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 in moving the box?
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 in moving the box?
d = 6.00 m
F = 50.0 N
ΔPE = 250. J
W = Fd = 50.0 N(6 m)
= 300. J
W = ΔPE + Wf
Wf = 50.0 J
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
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?
| 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?
| 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
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
Static Electricity