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
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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1 Answer
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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 downward, hence the work done is negative.
(c) The direction of motion of the object is in the opposite direction of the frictional force; hence the work done is negative.
(d) While a body moves on a rough horizontal plane, the frictional forces try to oppose the motion. But since the applied force maintains uniform velocity
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Using Newton’s formula,
This is a multiple choice answer as classified in NCERT Exemplar

This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”

Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1= ………….2
From eqn 1 and 2
So v1 = v’/ =v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path
This is a multiple choice answer as classified in NCERT Exemplar
(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 =-mgL + mgL=0
As the point of application of the contact forces does not move hence work done by reaction forces will be zero.
This is a multiple choice answer as classified in NCERT Exemplar
(c) m =150g =3/20kg
Time of contact =0.001s
U=126km/h=
V= -35m/s
Change in momentum of the ball = m (v-u)=
=21/2
F= dp/dt=- = - 1.05
Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.
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