Work, Energy and Power

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New answer posted•a year ago

Work, Energy and Power Class 11th

Why is there a 1/2 in the formula of kinetic energy?

2 Views|Posted a year ago

Asked by Aadit Singh Uppal

Answered by

Aadit Singh Uppal | Contributor-Level 10

a year ago

The 1/2 is a result of mathematical calculation, which occurs when we integrate? vdv in the formula of work done according to Newton's second law of motion. Without this, the final result will turn out to be twice of the actual value.

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New answer posted•a year ago

Work, Energy and Power Class 11th

A porter lifts a heavy suitcase of mass 80kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase.

(take g = 9.8 ms^-²)

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

W = mgh

$= 80 \times 9.8 \times (\frac{80}{100})$

$= 627.2 J$

$\begin{array}{l} \Rightarrow w_{f} + w_{g} = 0 \\ \Rightarrow w_{f} = - 627.2 J \end{array}$

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New answer posted•a year ago

Work, Energy and Power Class 11th

In a spring gun having spring constant 100N/m a small ball ‘B’ of mass 100g is put in its barrel (as shown in figure) by compressing the spring through 0.05m. There should be a box placed at a distance ‘d’ on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2m above the ground. The value of d is__________m.

15 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Let ball starts its motion with horizontal velocity v₀, so with the help of conservation of mechanical energy, we can write

$\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m x^{2} \Rightarrow v_{0} = x \sqrt[]{\frac{k}{m}} = 0.05 \times \sqrt[]{\frac{100}{0.1}} = 0.5 \times \sqrt[]{10} m / s$

t = Time required to fall the ball = $\sqrt[]{\frac{2 h}{g}} = 2 \times \sqrt[]{\frac{1}{10}} s$

^{$= d = v_{0} \times t = 0.5 \times \sqrt[]{10} \times 2 \times \sqrt[]{\frac{1}{10}} = 1 m$} &nbs

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New answer posted•a year ago

Work, Energy and Power Class 11th

A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time ‘t’ is proportional to:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

According to question, we can write

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New answer posted•a year ago

Work, Energy and Power Class 11th

If the kinetic energy of a moving body becomes four times its initial kinetic energy, then the percentage change in its momentum will be:

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$p = \sqrt[]{2 m K} \Rightarrow \frac{p_{2}}{p_{1}} = \sqrt[]{\frac{K_{2}}{K_{1}}} \Rightarrow \frac{p_{2}}{p} = \sqrt[]{\frac{4 K_{1}}{K_{1}}} = 2 \Rightarrow p_{2} = 2 p$

The percentage change in its momentum = $\frac{Δ p}{p} \times 100 = \frac{p}{p} \times 100 = 100 %$

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New answer posted•a year ago

Work, Energy and Power Class 11th

A mass of 5kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?

49 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Maximum energy = 10 J

$\frac{1}{2} K x^{2} = 10$

K = 5

Given T_pendulum = T_spring

$2 π \sqrt[]{\frac{l}{g}} = 2 π \sqrt[]{\frac{m}{K}}$

$\sqrt[]{\frac{4}{g}} = \sqrt[]{\frac{5}{5}}$

g = 4m/s²

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New answer posted•a year ago

Work, Energy and Power Class 11th

Due to cold weather a 1m water pipe of cross-sectional area 1 cm² is filled with ice at -10°C. Resistivity heating is used to melt the ice. Current of 0.5A is passed through 4 k $Ω$  resistance. Assuming that all the heat produced is used for melting, what is the minimum time required?

(Given latent heat of fusion for water/ice = 3.33 × 10⁵ J kg^-¹, specific heat of ice = 2 × 10³ J kg^-1 and density of ice = 10³ kg/m³)

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Energy required to melt

Q = $M S Δ T + M L$

$\Rightarrow 1 0^{- 1} \times 2 \times 1 0^{3} \times 10 + 1 0^{- 1} \times 3.33 \times 1 0^{5}$

->3.53 × 10⁴ J

Heat produce in wire

H = l²RT

$Q = 3.53 \times 1 0^{4} = {(\frac{1}{2})}^{2} \times (4 \times 1 0^{3}) \times t$

$t = \frac{3.53 \times 1 0^{4} \times 4}{4 \times 1 0^{3}} = 35.3 s e c$

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New answer posted•a year ago

Work, Energy and Power Class 11th

An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 40,000kg is moving with a speed of 75 kmh^-1 when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0m. If 90% of energy of the wagon is lost due to friction, the spring constant is________× 10⁵ N/m.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$= 0 - \frac{1}{2} m V^{2}$

As 90% of K.E. is losed by friction so that

$- \frac{90}{100} (\frac{1}{2} m V^{2}) - \frac{1}{2} K x^{2} = - \frac{1}{2} m V^{2}$

-K -> -16 × 10⁵

K = 16 × 10⁵

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New answer posted•a year ago

Work, Energy and Power Class 11th

An object of mass ‘m’ is being moved with a constant velocity under the actin of an applied force of 2N along a frictionless surface with following surface profile.

The correct applied force vs distance graph will be:

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Object is moving in upward direction with constant velocity so in upward motion (+2N) and for downward motion (-2N) So option (1) is correct representation.

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New answer posted•a year ago

Work, Energy and Power Class 11th