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

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9 months ago

6.1 (a) While the person lifts a bucket out of a well by means of a rope tied to the bucket, the direction of both the force and the displacement are same, hence the work done is positive.

(b) While lifting the bucket, he works against gravity, but the work done by the gravitational force is downw

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Using Newton’s formula,

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Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. 6.10. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.

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(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.

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This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.

a) So spring will continue compress until the blocks moves with the same velocity.

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(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv’

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-