Conservation of Mechanical Energy: Simple Pendulums and Non-Conservative Forces, Summaries of Acting

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Conservation of

mechanical energy

Lecture 11

Pre-reading: KJF §10.

2

Conservation of Mechanical

Energy

Under the influence of conservative forces only

(i.e. no friction or drag etc.)

M.E. = K + U = constant

Note that U and K can include such things as elastic

potential energy, rotational kinetic energy, etc.

Example : simple pendulum or slippery dip

(if friction & air resistance are negligible).

KJF §10.

Example: Tarzan

Tarzan who weighs 688N swings from a cliff at the

end of a convenient vine that is 18m long. From the

top of the cliff to the bottom of the swing he descends

by 3.2m.

(a) What is his speed at the bottom of the swing?

Neglect air resistance.

(b) The vine will break if the force on the vine

exceeds 950N. Does it break at the bottom of

the swing?

[7.9 m.s–1, no]

Loop the loop

What height does the ball have to start at to make

it through the loop?

Example 1 :

Block on horizontal surface slides to rest due to kinetic

friction. Work done by friction is

∆ME = ∆ K = – F k d

Example 2 :

Block sliding along a horizontal surface at constant

velocity. If work is done AGAINST friction by an applied

force F app and Δ K & Δ U = 0 then;

The amount of thermal energy produced must be exactly

equal to the amount of work done, in other words…

Work and Friction (1)

Work and Friction (2)

W = F s cosθ

Force working AGAINST friction is F appl = – F k

(why?) but F k = μk F N , cosθ = 1, and s = d , so the amount

of thermal energy produced is

∆ E th = F k d = μk F N d

Clearly, here work is not reversible. (Why not?)

Work done BY friction - same magnitude, opposite sign

A 20 kg child, starting from rest,

slides down a 3m high frictionless

slide.

How fast is he going at the

bottom?

[7.7 ms

• ]

Now he slides down a slide with friction, and his

speed at the bottom is 6.0 ms

• . How much

thermal energy has been produced by friction?

[228 J]

Problem: Skier

A 60 kg skier leaves the end of a ski jump ramp with a

velocity of 24 ms

• directed 25° above the horizontal.

Suppose that as a result of air resistance the skier returns to

the ground with a speed of 22 ms

• and lands at a point down

the hill that is 14m below the ramp.

How much energy is dissipated by air resistance during the

jump?

[11 kJ]

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Energy ILD