Question

k.e=1\2 mv2 proof definition of specific heat

Answer 1

HELLO THERE!

(i) Proof of Kinetic Energy = 1/2 mv²

Method 1: Consider a body starts from rest, and is accelerated to velocity v with acceleration a in moving a distance S.

Then, from Newton's Second Law of Motion,

F = ma.

And, from the equation of motion,

v² = u² + 2aS, since u = 0, v² = 2aS

So,

Work Done = F . S

$= ma \times \frac{v^{2}}{2a} \\\\= \frac{mv^{2}}{2} \\\\= \frac{1}{2} mv^{2}$

Here, work done is equal to the increase in Kinetic energy, so

K.E = 1/2 mv²

Method 2 (Using Calculus): When the force acting on the body of mass m is variable in magnitude, the work done by the force in moving the body through an infinitesimally small distance dS is:

dW = F.dS (Since for the infinitesimally small time interval, the Force is constant).

If a is the acceleration produced in the body,

F = ma.

So,

dW = ma dS

$\implies dW = m\frac{dv}{dt}dS \\\\\implies dW = m\frac{dS}{dt}dv \\\\\implies dW = mv dv$

[Since, a = dv/dt, v = dS/dt].

Now, integrate both sides with limits. Limits of left hand side = 0 to W and right hand side = 0 to v.

$dW = mv dv \\\\\implies\int\limits^W_0{dW} = \int\limits^v_0 {mv} \, dv \\\\\implies W = m\int\limits^v_0 {v} \, dv \\\\\implies W = m[\frac{v^{2}}{2}]^{v}_{0} \\\\\implies W = \frac{mv^{2}}{2} \\\\or, W = \frac{1}{2}mv^{2}$

Hence, K.E. = 1/2 mv²

(ii) Definition of Specific Heat: Specific heat is the amount of heat per unit mass required to raise the temperature of the body by one degree Celsius.

HOPE THIS HELPS.

Thanks!