Work Energy Theorem

Dear students in 9th and 10th standard we have learnt about the energy and law of conservation of energy. According to law of conservation of energy,’ Energy can neither be created nor be destroyed can be converted from one form to another form’.

Some of the conditions where this can be easily observed are, 1) water stored in the dam has potential energy and gets converted in to kinetic energy when windows of the dam are opened. 2) Children in merry go round looses potential energy while coming from the top of round to its bottom, whereas kinetic energy increases in same direction.

Work is nothing but the form of energy. Hence the change in energy can gives rise the work and same we are going to learn in the given article, so let’s learn it!

Let’s learn the vector product of vectors in details….!

F = mass.acceleration

∴ F = m(-a)

Negative sign indicates that the displacement is opposite to force

∴ a = -F/m      …………(1)

Using third equation of motion we can write

∴v2 = u2 + 2 as

∴ v2 – u2 =2 as

∴ v2 – u2 =2 (-F/m)s

∴ m (v2 – u2) = -2 F.s

∴ mu2 – mv2 = 2 F.s

Dividing both sides by 2 we can get,

∴ 1/2 mu2 – 1/2  mv2 = F.s

∴ 1/2 mu2 – 1/2  mv2 = W       (F.s =work done)

Left-hand side of above equation indicates the decrease in the kinetic energy and the right-hand side is the work done by the force. Thus, change in kinetic energy is equal to work done by the force, which is nothing but work-energy theorem.

Important condition:

Suppose object dropped from height h falls through air. During this its potential energy decreases and work is also done against the air resistance or frictional force. Hence the total work is not entirely converted into kinetic energy but some part of it is used in overcoming the air resistance and causes heat, sound etc.

PE + KE + Wair resistance

Let’s learn some numerical on vector product of vectors…!

1.) Force applied on object of mass 100 g changes its velocity from 54 kmph to 108 kmph. Find the work done in the process using work energy theorem.

Solution:

mass, m =100 g = 0.1 kg

u=54 kmph = 15 m/s, u=108 kmph = 30 m/s,

∴ W =1/2 mv2 – 1/2  mu2

∴ W = 1/2 m (v2 – u2)

∴ W = ½ × 0.1 × (302 – 152)

∴W = ½ × 0.1 × (900 – 225)

∴ W = 33.75 J

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