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

8 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 $θ = 1 * 10 * 10 s i n 30$ = 50J

(b) work done against friction = fs

= $μ N (s) = μ m g c o s θ$

= 0.1 $* 1 * 10 * c o s 30 * 10 = 8.66 J$

(c) increase in PE =mgh

=1 $* 10 * 10 * s i n 30$

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

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

= -mgh-fs+FS

= -50-8.6

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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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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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Payal Gupta

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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

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

So v₁ = v’/

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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 =-

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 $\times 10^{4} N$

Here – negative sign indicates that force will be opposite to the direction of movement of the ball be

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