ENERGY PROBLEMS

Superfluids

The strange behavior of liquid helium

BBC

ENERGY PROBLEMS

When work is done on
or by a system

Ex 1) An object slides across a horizontal table which changes its KE 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?

W_f = 2 J

**Work against friction increases
an object's Internal Energy** (makes it hotter)

Friction is a
nonconservative force
(doesn't store energy)

(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 + W_f

W_f - work done against friction

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 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 = (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.

What is the objects new P.E.?

Ex 3) 100 Joules of work is needed to lift a 100 kg object from the ground.

What is the objects new P.E. ?

W = ΔPE = 100J

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 + W_f

300. J = 250. J + W_f

W_f = 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 PE?

What is the mass of the object?

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?

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?

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?

k = 2.0 N/m

x = 5.0 m

W = ?

W = PE = 1/2kx²

= (1/2)2.0 N/m(5.0 m)²

= 25. J

Static Electricity