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New answer posted•a year ago

Work, Energy and Power Class 11th

A pendulum of length 2m consists of a wooden bob of mass 50g. A bullet of mass 75g is fired towards the stationary bob with a speed $υ .$  The bullet emerges out of the bob with a speed $\frac{υ}{3}$  and the bob just completes the vertical circle. The value of $υ$  is_________ms^-¹.

(if g = 10 m/s²).

13 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Com

m1v + m2 × 0 = m1v1 + m2v2

75v = $45 \frac{v}{3} + 50 \times 10$ $\frac{}{}$

$\Rightarrow 75 \times \frac{2 v}{3} = 50 \times 10$

$\Rightarrow v = \frac{3 \times 50 \times 10}{75 \times 2} = 10 m / s e c$

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New answer posted•a year ago

Work, Energy and Power Class 11th

What percentage of kinetic energy of a moving particle is transferred to a stationary particl when it strikes the stationary particle of 5 times its mass?

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Com (Conservation of Momentum)

Pi = Pf

$\Rightarrow m u_{1} + 5 m u_{2} = m v_{1} + 5 m v_{2}$

$\Rightarrow m u_{1} = m v_{1} + 5 m v_{2}$

$\Rightarrow u_{1} + v_{1} + 5 v_{2}$ …….(i)

& $e = 1 = \frac{v_{2} - v_{1}}{U_{1} - U_{2}} = \frac{v_{2} - v_{1}}{u_{1}}$ $\frac{_{}_{}}{_{}_{}} \frac{_{}_{}}{_{}}$

$_{}_{}_{}$ $\Rightarrow u_{1} = v_{2} - v_{1}$ …….(ii)

Equation (i) + (ii)

$k . E_{T} = \frac{5}{9} {(K . E)}_{i} = 55.6 %$

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New answer posted•a year ago

Work, Energy and Power Class 11th

An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being $68 k g$ . The mass of the elevator itself is $920 k g$ and it moves with a constant speed of $3 m / s$ . the frictional force opposing the motion is $6000 N$ . If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator $(g = 10 m / s^{2})$ must be at least:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Net force on motor will be

$F_{min} = [920 + 68 (10)] g + 6000$

$= 22000 N$

So, required power for motor

$\begin{array}{r} P_{min} & = {\overset{?}{F}}_{min} \cdot \overset{?}{v} \\ = 22000 \times 3 \\ = 66000 watt \end{array}$

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New answer posted•a year ago

Work, Energy and Power Class 11th

A body of mass 0.5 kg travels on straight line path velocity u = $(3 x^{3} + 4)$ $^{}$ m/s. The net work done by the force during its displacement from x = 0 to x = 2m is:

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$v_{i} = 3 \times {(0)}^{2} + 4 = 4 m / s$

$v_{F} = 3 \times 2^{2} + 4 = 16 m / s$

$ω = Δ k = \frac{1}{2} m [v_{F}^{2} - v_{i}^{2}]$

= $\frac{1}{2} \times 0.5 [1 6^{2} - 4^{2}] = \frac{240}{4} = 60 J$

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New answer posted•a year ago

Work, Energy and Power Class 11th

A bag of sand of mass 9.8kg is suspended by a rope. A bullet of 200g travelling with speed 10 ms^-1 gets embedded in it, then loss of kinetic energy will be

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$v = \frac{m v_{0}}{M + m} \Rightarrow K_{f} = \frac{1}{2} (M + m) v^{2} = \frac{m v_{0}^{2}}{2 (M + m)}$

$K_{i} = \frac{m v_{0}^{2}}{2}$

Loss in Kinetic energy = $Δ K = K_{i} - K_{f}$

$Δ K = \frac{1}{2} m v_{0}^{2} (\frac{M}{M + m}) = \frac{1}{2} \times 0.2 \times 100 = [\frac{9.8}{10}] = 9.8 J$

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New answer posted•a year ago

Work, Energy and Power Class 11th

If momentum of a body is increased by 20%, then its kinetic energy increases by

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\frac{Δ K}{K_{i}} \times 100 = \frac{K_{f} - K_{i}}{K_{i}} \times 100$

$= \frac{\frac{P_{f}^{2}}{2 m} - \frac{P_{i}^{2}}{2 m}}{\frac{P_{i}^{2}}{2 m}} \times 100 = \frac{P_{f}^{2} - P_{i}^{2}}{P_{i}^{2}} \times 100$

$= \frac{{(1.2 P_{i})}^{2} - {(P_{i})}^{2}}{P_{i}^{2}} \times 100 = \frac{1.44 P_{i}^{2} - P_{i}^{2}}{P_{i}^{2}} \times 100$

= 44%

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New answer posted•a year ago

Work, Energy and Power Class 11th

A particle $(m = 1 k g)$ slides down a frictionless track (AOC) starting from rest of a point $A$ (height $2 m$ ). After reaching $C$ , the particle continues to move freely in air as a projectile. When it leading its highest point $P$ (height 1 m ) the kinetic energy of the particle (in J) is: (Figure drawn is schematic and not to scale (take $g = 10 {m s}^{- 2}$ )

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$K E = {P E}_{1} - {P E}_{2} = m g h_{1} - {m g h}_{2}$

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New answer posted•a year ago

Work, Energy and Power Class 11th

A block of mass ‘m’ (as shown in figure) moving with kinetic energy E compresses a spring through a distance 25 cm when, its speed is halved. The value of spring constant of used spring will be nE Nm^-¹ for n = _________.

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

loss in k = Gain in U spring

$\Rightarrow \frac{3 E}{4} = \frac{1}{2} k {(\frac{25}{100})}^{2}$

$\Rightarrow \frac{3 E}{4} = \frac{k}{32} \Rightarrow k = \frac{3 F}{4} \times 32 = 24 E$

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New answer posted•a year ago

Work, Energy and Power Class 11th

A 0.5kg block moving at a speed of 12ms^-¹ compresses a spring through a distance 30cm when its speed is halved. The spring constant of the spring will be…………Nm^-¹.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

A particle experiences a variable for $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$  in a horizontal x-y plane. Assume distance in meters and force is Newton. if the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then kinetic Energy changes by :

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\vec{F} = 4 x \hat{i} + 3 y^{2} \hat{j}$

$\begin{array}{l} d \vec{r} = d \vec{s} = d x \hat{i} + d y \hat{j} \\ \int d w = \int \vec{F} . d \vec{r} \\ = \int (4 x \hat{i} + 3 y^{2} \hat{j}) . (d x \hat{i} + d y \hat{j}) \\ = \int_{1}^{2} 4 x d x + \int_{2}^{3} 3 y^{2} d y \\ = \frac{4}{2} {[x^{2}]}_{1}^{2} + \frac{3}{3} {[y^{3}]}_{2}^{3} \\ = 2 [4 - 1] + [27 - 8] \end{array}$

= 6 + 19 = 25 J

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