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Work, Energy, and Power physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface in fig

The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate

(a) work done against gravity

(b) work done against force of friction

(c) increase in potential energy

(d) increase in kinetic energy

(e) work done by applied force.

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Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

(a) work done= increase in PE

= mg (vertical distance travelled)

= mg (s)sin $θ = 1 \times 10 \times 10 s i n 30$ = 50J

(b) work done against friction = fs

= $μ N (s) = μ m g c o s θ$

= 0.1 $\times 1 \times 10 \times c o s 30 \times 10 = 8.66 J$

(c) increase in PE =mgh

=1 $\times 10 \times 10 \times s i n 30$

$= 100 (\frac{1}{2}) = 50 J$

(d) according to work energy theorem W= change in KE

= -mgh-fs+FS

= -50-8.66+10 (10)

= 41.34J

(e) force f = FS

= 10 (10)= 100J

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Using Newton’s formula,

$\Rightarrow (f + d_{1}) (f - d_{2}) = f^{2}$

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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.

(c) If spring is massless, the final state of the M₁ is state of rest.

(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

Payal Gupta

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.

b) As the surfaces are frictionless momentum of the system will conserved.

c) If the spring is massless whole energy of M1 will be imparted and M1 will be rest .

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A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

(a) The velocity of the bullet will be reduced to half its initial value.

(b) The velocity of the bullet will be more than half of its earlier velocity.

(c) The bullet will continue to move along the same parabolic path.

(d) The bullet will move in a different parabolic path.

(e) The bullet will fall vertically downward after hitting the target.

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Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(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

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

So v₁ = v’/ $\sqrt[]{2}$ =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

...more

This is a multiple choice answer as classified in NCERT Exemplar

(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

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

So v₁ = v’/ $\sqrt[]{2}$ =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, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.

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A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.

(a) Work done by all forces on man is equal to the rise in potential energy mgL.

(b) Work done by all forces on man is zero.

(c) Work done by the gravitational force on man is mgL.

(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

Payal Gupta

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.

A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be

(a) 10.5 N

(b) 21 N

(c) 1.05 ×104 N

(d) 2.1 × 104 N

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

Time of contact =0.001s

U=126km/h= $\frac{126 \times 1000}{60 \times 60} = \frac{35 m}{s}$

V= -35m/s

Change in momentum of the ball = m (v-u)= $\frac{3}{20} (- 35 - 35) k g m / s$

=21/2

F= dp/dt=- $\frac{21 / 2}{0.001} N$ = - 1.05 $\times 10^{4} N$

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