# NCERT Class 11 Physics Solutions: Chapter 6 – Work, Energy, and Power Part 2

Get unlimited access to the best preparation resource for NSO Class-11: fully solved questions with step-by-step explanation- practice your way to success.

**Question 6.3:**

Given in Figure are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

(a) (b)

(c) (d)

**Answer:**

(a)

Total energy of a system is given by the relation:

Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where . becomes negative. In the given case, the potential energy of the particle becomes greater than total energy . Hence, kinetic energy becomes negative in this region. Therefore, the particle will not exist is this region. The minimum total energy of the particle is zero.

(b) All regions

In the given case, the potential energy ) is greater than total energy in all regions. Hence, the particle will not exist in this region.

(c)

In the given case, the condition regarding the positivity of is satisfied only in the region between and

The minimum potential energy in this case is. Therefore, . Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than . So, the minimum total energy the particle must have is .

(d)

In the given case, the potential energy) of the particle becomes greater than the total energy for . Therefore, the particle will not exist in these regions.

The minimum potential energy in this case is Therefore. Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than. So, the minimum total energy the particle must have is.

**Question 6.4:**

The potential energy function for a particle executing linear simple harmonic motion is given by where k is the force constant of the oscillator. For , the graph of versus is shown in Figure. Show that a particle of total energy moving under this potential must ‘turn back’ when it reaches

**Answer :**

Total energy of the particle,

Force constant,

Kinetic energy of the particle,

According to the conservation law:

Now of ‘turn back’, velocity (and hence K) becomes zero.

Hence, the particle turns back when it reaches