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6 months ago

Work, Energy and Power Class 11th

6.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given $θ_{1}$ = 30 $°$ , $θ_{2}$ = 60 $°$ , and h = 10 m, what are the speeds and times taken by the two stones ?

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Payal Gupta

Contributor-Level 10

6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ &

...more

6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ = v/ $a_{1}$ = v/g $\sin ? ?_{1}$ = 14/(9.81 $\sin ? 30 °)$ = 2.85 s

$t_{2}$ = v/ $a_{2}$ = v/g $\sin ? ?_{2}$ = 14/(9.81 $\sin ? 60 °)$ = 1.65 s

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New answer posted

6 months ago

Work, Energy and Power Class 11th

6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

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Payal Gupta

Contributor-Level 10

6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 0.4 kg

Initial speed of the wooden block = 0

Let's assume, final speed of the bullet = v

Applying the law of conservation of momentum

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2

...more

6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 0.4 kg

Initial speed of the wooden block = 0

Let's assume, final speed of the bullet = v

Applying the law of conservation of momentum

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2) $*$ 0.012 X $70^{2}$ - (1/2) $* (0.012 + 0.4) * {2.04}^{2}$ J

= 29.4 – 0.857 J

= 28.54 J

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6 months ago

Work, Energy and Power Class 11th

6.23 A family uses 8 kW of power.

(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

(b) Compare this area to that of the roof of a typical house.

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Payal Gupta

Contributor-Level 10

6.23 Power used by the family = 8 kW = 8000 W

(a) Solar energy received = 200 W/ $m^{2}$

Percentage conversion of Solar energy to Electrical energy = 20%

If the area required is A then 0.2 $* A * 200 = 8000$

A = 200 $m^{2}$ . The comparable roof size is 14.14 X 14.14 m

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6 months ago

Work, Energy and Power Class 11th

6.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

(a) How much work does she do against the gravitational force ?

(b) Fat supplies 3.8 * 10⁷J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

6.22 Mass lifted, m = 10 kg

Height to which the mass lifted, h = 0.5 m

No of repetitions, n = 1000

(a) Work done against gravitational force,

W = nmgh = 1000 $* 10 * 9.81 * 0.5 =$ 49050 J

(b) Mechanical energy supplied by 1 kg fat, with 20% efficiency rate = 0.2 $*$ 3.8 $* 10^{7}$ = 0.76 $* 10^{7}$ J/kg

Fat used by dieter = 49050 / (0.76 $* 10^{7})$ kg = $6.45 * 10^{- 3}$ kg

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New answer posted

6 months ago

Work, Energy and Power Class 11th

6.21 The blades of a windmill sweep out a circle of area A.

(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?

(b) What is the kinetic energy of the air ?

(c) Assume that the windmill converts 25% of the wind's energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ?

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Payal Gupta

Contributor-Level 10

6.21 Given, the area of the windmill sweep = A, Wind velocity = v

The volume of air passing through the blade = Av

Let the density of air be $?$ , the mass of air passing through the blade = $? A v$

(a) The mass of air passing through the blade in time t = $? A v t$

(b) The kinetic energy of air = $(\frac{1}{2}) m v^{2}$ = $(\frac{1}{2}) ? A v t v^{2}$ = $(? A t v^{3})$ /2 …. (1)

(c) Area, A = 30 $m^{2}$ , v = 36 km/h = 10 m/s, density of air be $?$ = 1.2 kg/ $m^{3}$

Total wind energy, from eqn. (1) = 18 kW

Electrical energy = 25 % of wind energy = 0.25 $* 18 = 4.5 k W$

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New answer posted

6 months ago

Work, Energy and Power Class 11th

6.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x ^3/2 where a = 5 m^–1/2s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

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Payal Gupta

Contributor-Level 10

6.20 Mass of the body = 0.5 kg

Velocity, v = a x 3/2

a = 5 m–1/2 s–1

At x =0, the initial velocity, u = 0

At x = 2, the final velocity, v = 5 $* 2^{3 / 2}$ = 14.142 m/s

Work done by the system = increase in K.E. of the body = (1/2)m ( $v^{2}$ - $u^{2}$ )

= (1/2) $* 0.5 *$ 14.142 $* 14.142$ = 50 J

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6 months ago

Work, Energy and Power Class 11th

6.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty?

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Payal Gupta

Contributor-Level 10

6.19 Given, the mass of the trolley, $m_{t}$ = 300 kg, mass of the sand bag, $m_{s}$ = 25 kg, uniform velocity of the trolley, v = 27 kmph = 0.75 m/s

Since there is no external force acting on the system, the speed of the trolley will remain unchanged even after entire sand is empty. 27 kmph is the answer.

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6 months ago

Work, Energy and Power Class 11th

6.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

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Payal Gupta

Contributor-Level 10

6.18 The length of the pendulum, l = 1.5 m

The potential energy of the bob at horizontal position = mgl

Since it dissipates 5% of its kinetic energy to come to the horizontal position, from the law of conservation of energy we get,

$(\frac{1}{2}) m v^{2}$ = (0.95) $* m g l$

$v^{2}$ = 2 $* 0.95 * 9.81 * 1.5$

v = 5.287 m/s

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6 months ago

Work, Energy and Power Class 11th

6.17 The bob A of a pendulum released from 30 $°$ to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.

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Payal Gupta

Contributor-Level 10

6.17 In an elastic collision, when the ball A will hit the ball B, A comes to rest immediately and the ball B acquires the velocity of ball A. The momentum thus gets transferred from a moving body to a stationary body.

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6 months ago

Work, Energy and Power Class 11th

6.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision ?

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Payal Gupta

Contributor-Level 10

6.16 The mass of the ball bearing = m

Before the collision, the total K.E. of the system = K.E. of the stationary ball bearing + K.E. of the striking ball bearing = 0 + $(\frac{1}{2}) {m v}^{2}$

After the collision, the K.E. of the total system is

(a) Case (i) = 0 + $(\frac{1}{2}) (2 m) {(\frac{v}{2})}^{2}$ = $(\frac{1}{4}) {m v}^{2}$

(b) Case (ii) = 0 + $(\frac{1}{2}) m v^{2}$

(c) Case (iii) = $(\frac{1}{2}) (3 m) {(\frac{v}{3}) /}^{2}$ = $(\frac{1}{6}) m v^{2}$

Case (ii) is possible since K.E. is conserved in this case.

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