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11th Standard English Medium Physics Subject Work, Energy and Power Creative 3 Mark Questions with Solution Part - II

Published on: 24/06/2021

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Std: 11th Standard - Physics

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Questions:

Generic 3

5 x 3 = 15

One coolie takes 1 min to raise a box through a height 2m. Another takes 30 m/s for the same job and does the same amount of work. Which one of these two has a greater power?

The potential energy of a spring when stretched through a distance x is 10J. What is the amount of work done on the same spring to sketch it through an additional distance 'x'?

Calculate the velocity of the bob of a simple pendulum at its mean position if it is able to rise to a vertical height of 10 cm. Take g = 9.8 m/s².

A partice moves along the x-axis from x = 0 to x = 5m under the influence of a force given by F = 7 - 2x + 3x². Find the work done in the process.

What is the stopping distance for a vehicle, of mass m moving with speed v along a level road, if the co-efficient of friction between the tyres and the road is \(\mu\)?

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Answers Key:

Generic 3

5 x 3 = 15

Both the coolies do the same amount of work, so they spend the same amount of energy.
W =P₁t₁ = P₂t₂
\({p_2 \over p_1}={t_1 \over t_2}={60\ sec \over 30\ sec}=2\)
(or) P₂ = 2P₁ 2^nd coolie has greater power.

\(U={1 \over 2}{kx}^{2}=10J\)
When x becomes 2x, the P.E will be, \(U'={1\over 2}k(2x)^2=4\left({1 \over 2}{kx}^{2} \right)\)
= 4\(\times\)10 = 40J
\(\therefore\) Work done = V' - v
=40-10 = 30 J

\(\begin{matrix} K.E.of\ the\ bob \\ at\ Its\ mean\ position \end{matrix}\} =P.E.\ at\ its\ highest\ position\)
\({1\over 2}{mv}^{2}=mgh\)
\(v=\sqrt{2gh}\)
\(=\sqrt{2\times9.8\times10}=\sqrt{1.96}\)
v = 1.4 m/s

Work done ill moving the particle from x = 0 to x = 5m will be
\(W=\int_{0}^{5}Fdt=\int_{0}^{5}(7-2x+3x^2)dx \)
\(={\left[ 7x=-{2x^2\over2}+{3x^2\over 3} \right]}^{5}={[7x-x^2+x^3]}_{0}^{5}=35-25+125=135J\)

When the vehicle of mass m is moving with velocity v, the kinetic energy of the vehicle \(k=\left( {1\over 2} \right){mv}^{2}\) and if s is the stopping distance, then work done by friction.
W = \(fs\ \cos\theta=\mu mg\ s\ \cos 180°\)
So by work - Energy theorem, \(W=\triangle k = k_f-k_i\)
i.e, \(-\mu mgs=0-{1\over 2}{mv}^{2}\) or \(s={v^2\over 2\mu g}\)