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INTRODUCTION - Chapter 6 Work Energy & Power class 11 ncert solutions Physics - SaraNextGen [2024]


Question 6.1:

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.

(b) work done by gravitational force in the above case,

(c) work done by friction on a body sliding down an inclined plane,

(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,

(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:

(a) Positive

In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.

(b) Negative

In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.

(c) Negative

Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.

(d) Positive

Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.

(e) Negative

The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

Question 6.2:

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.

Compute the

(a) work done by the applied force in 10 s,

(b) work done by friction in 10 s,

(c) work done by the net force on the body in 10 s,

(d) change in kinetic energy of the body in 10 s,

and interpret your results.

Answer:

Mass of the body, m = 2 kg

Applied force, F = 7 N

Coefficient of kinetic friction, µ = 0.1

Initial velocity, u = 0

Time, t = 10 s

The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4474/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_3c132763.gif

Frictional force is given as:

f = µmg

= 0.1 × 2 × 9.8 = – 1.96 N

The acceleration produced by the frictional force:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4474/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_m4dc80009.gif https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4474/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_m7f772b4a.gif

Total acceleration of the body:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4474/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_m5b840268.gif

The distance travelled by the body is given by the equation of motion:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4474/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_1d68b3e1.gif

(a) Work done by the applied force, Wa = F × s = 7 × 126 = 882 J

(b) Work done by the frictional force, Wf = F × s = –1.96 × 126 = –247 J

(c) Net force = 7 + (–1.96) = 5.04 N

Work done by the net force, Wnet= 5.04 ×126 = 635 J

(d) From the first equation of motion, final velocity can be calculated as:

v = u + at

= 0 + 2.52 × 10 = 25.2 m/s

Change in kinetic energyhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4474/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_15e1db99.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4474/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_m7374da65.gif

Question 6.3:

Given in Fig. 6.11 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.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4476/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_m2bb43fec.jpg

Answer:

(a) x > a; 0

Total energy of a system is given by the relation:

E = P.E. + K. E.

∴K.E. = E – P.E.

Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where K.E. becomes negative.

In the given case, the potential energy (V0) of the particle becomes greater than total energy (E) for x > a. 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 (V0) is greater than total energy (E) in all regions. Hence, the particle will not exist in this region.

(c) x > a and x < b;  –V1

In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between > a and x < b.

The minimum potential energy in this case is –V1. Therfore, K.E.  = E – (–V1) = E + V1. Therefore, for the positivity of the kinetic energy, the totaol energy of the particle must be greater than –V1. So, the minimum total energy the particle must have is –V1.

(d) https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4476/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_m620364d1.gif

In the given case, the potential energy (V0) of the particle becomes greater than the total energy (E) forhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/184/4476/NS_20-10-08_Ravinder_11_Physics_6_30_NRJ_LVN_html_m32ff7a06.gif . Therefore, the particle will not exist in these regions.

The minimum potential energy in this case is –V1. Therfore, K.E.  = E – (–V1) = E + V1. Therefore, for the positivity of the kinetic energy, the totaol energy of the particle must be greater than –V1. So, the minimum total energy the particle must have is –V1.

Also Read : Page-No-135:-Chapter-6-Work-Energy-&-Power-class-11-ncert-solutions-Physics

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