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Science Que-Ans (Each of 5 Mark )

Work and Energy · MP Board (MPBSE) STD 9 (Madhya Pradesh - English Medium). 9 questions.

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Que-Ans (Each of 5 Mark )

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9 questions · timed · auto-graded

Question 15 Marks
A body of $2kg$ falls from rest. What will be its kinetic energy during the fall at the end of $2s? ($Assume $g = 10m/s^2)$
Answer
Mass of body $= 2kg$
Initial velocity $u = 0$
Time taken $= 2s$
Acceleration due to gravity, $g = 10m/s^2$ Final velocity $v$ Using first equation of motion $v = u + gt = 0 + 10 × 2 = 20m/s$ $\text{KE}=\frac{1}{2}\text{m}\times\text{v}^2$ $=\frac{1}{2}\times2\times20^2=400\text{J}$
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Question 25 Marks
State and explain the law of conservation of energy with an example.
Answer
Law of conservation of energy: Law of conservation of energy states that whenever energy changes from one form to another form, the total amount of energy remains constant. Energy can never be created nor destroyed, it transforms from one form to another. Example: When electrical energy is converted into light energy in an electric bulb, then some energy is wasted as heat during conversion but the total energy remains the same.
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Question 35 Marks
A ball of mass $200g$ falls from a height of $5$ metres. What is its kinetic energy when it just reaches the ground$? (8 = 9.8m/s$).
Answer
Mass $=200 \mathrm{~g}=0.2 \mathrm{~kg}$
Height $=5 \mathrm{~m}$
Initial velocity $\mathrm{u}=0$
Acceleration due to gravity, $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
Final velocity, $v$ Using third equation of motion, $v^2-u^2=2 g s $
$v^2-0=2 \times 9.8 \times 5 v^2=98 \ldots (i)$
Kinetic energy $=\frac{1}{2} m \times v^2$ Put the value of $\mathrm{v}^2$ from equ. $(i)$
Kinetic energy $=\frac{1}{2} \times 0.2 \times 98=9.8 \mathrm{~J}$
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Question 45 Marks
Two objects having equal masses are moving with uniform velocities of $2m/s$ and $6m/s$ respectively. Calculate the ratio of their kinetic energies.
Answer
Let masses of the two objects be $m.$
$\text{v}_1 = 2\text{m}/\text{s}$
$\text{v}_2 = 6\text{m}/\text{s}$
$\text{KE}_1=\frac{1}{2}\text{m}\times\text{v}_1^2$
$=\frac{1}{2}\text{m}\times2^2$
$\text{KE}_2=\frac{1}{2}\text{m}\times\text{v}_2^2$
$=\frac{1}{2}\text{m}\times6^2$
$\text{Ratio}=\frac{\text{KE}_1}{\text{KE}_2}$
$=\frac{2^2}{6^2}=\frac{4}{36}=\frac{1}{9}$
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Question 55 Marks
A bullet of mass $15g$ has a speed of $400m/s$. What is its kinetic energy? If the bullet strikes a thick target and is brought to rest in $2cm$, calculate the average net force acting on the bullet. What happens to the kinetic energy originally in the bullet?
Answer
Mass, $\mathrm{m}=15 \mathrm{~g}=0.015 \mathrm{~kg}$ Velocity, $\mathrm{v}=400 \mathrm{~m} / \mathrm{s} \mathrm{~KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \times 0.015 \times 400^2=1200 \mathrm{~J}$ Distance covered before stopping, $\mathrm{s}=2 \mathrm{~cm}=0.02 \mathrm{~cm}$ Here, initial velocity, $\mathrm{u}=400 \mathrm{~m} / \mathrm{s}$ Final velocity $=0 \mathrm{v}^2-\mathrm{u}^2=2 \mathrm{as} 0^2-400^2=2 \times \mathrm{a}$ $\times 0.02 \mathrm{a}=-4 \times 10^6 \mathrm{~m} / \mathrm{s}^2$ (Negative sign shows retardation) $\mathrm{F}=\mathrm{ma}=0.015 \times-4 \times 10^6=6 \times 10^4 \mathrm{~N}$ Kinetic energy of the bullet is mainly converted into heat energy (by friction).
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Question 65 Marks
Explain the transformation of energy in the following cases:
  1. A ball thrown upwards.
  2. A stone dropped from the roof of a building.
Answer
  1. When a ball is thrown upwards its kinetic energy gradually converts into potential energy and potential energy becomes maximum at the maximum height attained by the ball.
  2. When a stone is dropped from the roof of the building its potential energy gradually converts into kinetic energy and kinetic energy becomes maximum when the stone is just above the ground.
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Question 75 Marks
Calculate the work done by the brakes of a car of mass $1000kg$ when its speed is reduced from $20m/s$ to $10m/s?$
Answer
Mass of car $=1000 \mathrm{~kg}$
Initial velocity $\mathrm{u}=20 \mathrm{~m} / \mathrm{s}$
Final velocity $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$
Retardation $=\mathrm{a}$
Distance covered $=\mathrm{s}$
Using third equation of motion $\mathrm{v}^2-\mathrm{u}^2=2$ as $10^2-20^2=2$ 
as $=-\frac{300}{2} \ldots (i)$
as $=-150 \ldots (i)$
Work done $\mathrm{W}=\mathrm{F} \times \mathrm{s}$ But $F=m \times a$
So, $W=m \times a \times s$ Put the value of 'as' from equation $(i)$
$W=1000 \times-150=-150000=-150 \mathrm{~kJ}$
Negative sign implies that force of brakes acts opposite to the direction of motion.
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Question 85 Marks
The following data was obtained for a body of mass $1kg$ dropped from a height of $5$ metres:
 
Distance above ground
Velocity
$i.$
$5m$
$0m/s$
$ii.$
$3.2m$
$6m/s$
$iii.$
$0m$
$10m/s$
Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). $(g = 10m/s^2).$
Answer
Mass of body $= 1kg g = 10m/s^2 PE = m \times g \times h = 1 \times 10 \times h = 10 \times h$ $\text{KE}=\frac{1}{2}\text{mv}^2=\frac{1}{2}\times1\times\text{v}^2=\frac{\text{v}^2}{2}$
  1. At $h = 5m$
$v = 0m/s$
$PE = 10 \times 5 = 50J$
$KE = 0J$
Total energy $= PE + KE$$
  1. At $h = 3.2m$
$v = m/s$
$PE = 10 × 3.2 = 32J$
$\text{KE}=\frac{6^2}{2}=18\text{J}$
Total energy $= PE + KE$
$32 + 18 = 50J$
  1. At$ h = 0m$
$ v = 10m/s$
$PE = 0J$
$\text{KE}=\frac{10^2}{2}=50\text{J}$
Total energy = PE + KE = 50J
The total energy in all three cases is constant. This proves the law of conservation of energy.
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Question 95 Marks
On a level road, a scooterist applies brakes to slow down from a speed of $10m/s$ to $5m/s$. If the mass of the scooterist and the scooter be $150\ kg$, calculate the work done by the brakes. (Neglect air resistance and friction)
Answer
Mass of scooter $+$ scooterist $= 150kg$ Initial velocity $u = 10m/$s Final velocity $v = 5m/s$
Retardation $=$ a Distance covered $ = s$
Using third equation of motion $v^2 - u^2 = 2as 5^2 - 10^2 = 2as$
 $\text{as} =-\frac{75}{2}\ ...(\text{i})$
Work done $W = F \times s$ But $F = m \times a$
So, $W= m \times a \times s$ Put the value of ‘as’ from eq$(i)$
$\text{W} = 150\times\Big(-\frac{75}{2}\Big) = -5625 \text{J}$
Neagtive sign implies that force of brakes acts opposite to the direction of motion.
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