Work and Energy Test

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Work and Energy Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

A pile driver drives postes into the ground by repeatedly dropping a heavy object on them. Assume the object is dropped from the same height each time. By what factor does the energy of the pile driver-Earth system change when the mass of the object being dropped is doubled?

A
$\dfrac{1}{2}$

B
$1$ ; the energy is the sane

C
$2$

D
$4$

Solution
Question 2
1 / -0

A car weighing 600 kg traveling with $72\, km/h$ stops at a distance of $60\ m$ decelerating uniformly. What is the work done by brakes in joules?

A
$1.2 \times 10^5\ J$

B
$2.4 \times 10^5\ J$

C
$3.2 \times 10^5\ J$

D
$1.4 \times 10^5\ J$

Solution

Given,
Initial velocity, $u=72\ km/h$
$u=\dfrac{72\times 1000}{3600}\ m/s=20\ m/s$
Final velocity, $v=0\ m/s$
Displacement $s=60\ m$

Using Newton's third equation of motion,
$2as=v^2-u^2$

$\implies a=\dfrac{v^2-u^2}{2s}$

$\implies a=\dfrac{0^2-20^2}{2\times 60}$

$\implies a=\dfrac{0^2-20^2}{2\times 60}$

$\implies a=\dfrac{400}{120}$

$a= -\dfrac{10}{3}\ m/s^2$
Magnitude of acceleration, $a=\dfrac{-10}{3}\ m/s^2$

Force $F=ma=600\times (10/3)$
$F=2000\ N$

Work done, $W=F \times S$
$W=2000\times 60$
$W=120000 J$
Question 3
1 / -0

1 kWh is equal to

A
$3.6 \times 10^6 MJ$

B
$3.6 \times 10^5 MJ$

C
$3.6 \times 10^2 MJ$

D
$3.6 MJ$

Solution

1 kilowatt hour is the energy produced by 1 kilowatt power source in 1 hour.

$1kWh=1kW\times 1hour=1000\times 3600 W.s$

$\implies 1kWh=3.6\times 10^6J$

$\implies 1kWh=3.6MJ$

Answer-(D)
Question 4
1 / -0

A man carries a heavy box on his head on a horizontal plane from one place to another. In this case, he does

A
Maximum work

B
No work

C
Negative work

D
Minimum work

Solution

$\underline{\text{A man carrying a box on his head:}}$
In this case, the force acting on the man is the weight of the box, which is acting downwards.
The displacement of the man is along the horizontal direction.
Therefore, the force and the displacement are perpendicular to each other. So there is no work involved in this process.
Question 5
1 / -0

Energy cannot be measured in.

A
$Js^{-1}$

B
$Ws$

C
$kWh$

D
$erg$

Solution

Energy can be measured in Ws (i.e watt second), kWh and erg. $Js^{-1}$ is the unit of power $P=\dfrac{E}{t}$ , energy can not be measured in that.
Question 6
1 / -0

Sunil rolls a marble ball down a frictionless inclined plane as shown above. What kind of energies are possessed by the rolling marble ball?

A
(i) and (ii) only

B
(ii) and (iii) only

C
(iii) and (i) only

D
(i), (ii) and (iii)
Question 7
1 / -0

Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time as a function of time $t$ is given by

A
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }\quad$

B
$m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }$

C
$\cfrac { 1 }{ 2 } m{ \left( \cfrac { v }{ { t }_{ 1 } } \right) }^{ 2 }{ t }^{ 2 }$

D
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 }^{ 2 } } { t }^{ 2 }$

Solution

Ans : We know $W = F \times s$
where,
$F = ma$
By using Kinematics, $s = \dfrac{1}{2}at^{2}$ , if u =0

So,
$W = (ma).(\dfrac{1}{2}at^{2})$
$W = \dfrac{1}{2}ma^{2}t^{2}$ where $a = v/t_{1}$

$\boxed{W = \dfrac{1}{2}m(\dfrac{v}{t_{1}})^{2}t^{2}}$
Question 8
1 / -0

A man ${M}_{1}$ of mass $80kg$ runs up a staircase in $15s$ . Another man ${M}_{2}$ also of mass $80kg$ runs up the stair case in $20s$ . The ratio of the power developed by them will be:

A
$1$

B
$4/3$

C
$16/9$

D
None of the above

Solution

Let the height of the staircase is $h$ .

$\underline{\text{Power of the first man:}}$
Work done by the man = $mgh=80\times g\times h$
Time taken by the man to ascend the staircase is $15\ s$
Power of the man $P_1=\dfrac{mgh}{t}=\dfrac {80\times g\times h}{15}\ W$

$\underline{\text{Power of the second man:}}$
Work done by the man = $mgh=80\times g\times h$
Time taken by the man to ascend the staircase is $20\ s$
Power of the man $P_2=\dfrac{mgh}{t}=\dfrac {80\times g\times h}{20}\ W$

Thus, $\dfrac{P_1}{P_2}=\dfrac{20}{15}=\dfrac{4}{3}$
Question 9
1 / -0

A weight lifter lifts a weight off the ground and holds it up then:

A
Work is done in lifting as well as holding the weight

B
No work is done in both lifting and holding the weight

C
Work is done in lifting the weight but no work is required to hold it up

D
No work is done in lifting the weight but work is required to be done in holding it up

Solution

Work is done by the lifter in lifting the weight (displacement of weight upwards in the direction of force).
There is no displacement of weight when he holds it. Hence, no work is done.
Question 10
1 / -0

An athlete in the Olympic games covers a distance of $100$ m in $10$ s. His kinetic energy can be estimated to be in the range. (Assume weight = 60kg)

A
$200J-500J$

B
$2\times 10^5J-3\times 10^5J$

C
$20000J-50000J$

D
$2000J-5000J$

Solution

Velocity $V=\dfrac{distance}{time}=\dfrac{100}{10}=10 m/s$

Assuming his mass to be 60kg, his kinetic energy is
$K.E =\dfrac{1}{2}mv^2$

$= \dfrac{1}{2}\times 60\times 100$

$=60\times50$

$=3000 J$
Hence, Option D

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Re-Attempt Weekly Quiz Competition

Work and Energy Test - 52

Result

Work and Energy Test - 52

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SHARING IS CARING

Question 1

12\dfrac{1}{2}21​

111; the energy is the sane

222

444

Solution

Question 2

1.2×105 J1.2 \times 10^5\ J1.2×105 J

2.4×105 J2.4 \times 10^5\ J2.4×105 J

3.2×105 J3.2 \times 10^5\ J3.2×105 J

1.4×105 J1.4 \times 10^5\ J1.4×105 J

Solution

Question 3

3.6×106MJ3.6 \times 10^6 MJ3.6×106MJ

3.6×105MJ3.6 \times 10^5 MJ3.6×105MJ

3.6×102MJ3.6 \times 10^2 MJ3.6×102MJ

3.6MJ3.6 MJ3.6MJ

Solution

Question 4

Maximum work

No work

Negative work

Minimum work

Solution

Question 5

Js−1Js^{-1}Js−1

WsWsWs

kWhkWhkWh

ergergerg

Solution

Question 6

(i) and (ii) only

(ii) and (iii) only

(iii) and (i) only

(i), (ii) and (iii)

Question 7

12mvt1t2\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }\quad 21​mt1​v​t2

mvt1t2m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }mt1​v​t2

12m(vt1)2t2\cfrac { 1 }{ 2 } m{ \left( \cfrac { v }{ { t }_{ 1 } } \right) }^{ 2 }{ t }^{ 2 }21​m(t1​v​)2t2

12mvt12t2\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 }^{ 2 } } { t }^{ 2 }21​mt12​v​t2

Solution

Question 8

111

4/34/34/3

16/916/916/9

None of the above

Solution

Question 9

Work is done in lifting as well as holding the weight

No work is done in both lifting and holding the weight

Work is done in lifting the weight but no work is required to hold it up

No work is done in lifting the weight but work is required to be done in holding it up

Solution

Question 10

200J−500J200J-500J200J−500J

2×105J−3×105J2\times 10^5J-3\times 10^5J2×105J−3×105J

20000J−50000J20000J-50000J20000J−50000J

2000J−5000J2000J-5000J2000J−5000J

Solution

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$\dfrac{1}{2}$

$1$ ; the energy is the sane

$2$

$4$

$1.2 \times 10^5\ J$

$2.4 \times 10^5\ J$

$3.2 \times 10^5\ J$

$1.4 \times 10^5\ J$

$3.6 \times 10^6 MJ$

$3.6 \times 10^5 MJ$

$3.6 \times 10^2 MJ$

$3.6 MJ$

$Js^{-1}$

$Ws$

$kWh$

$erg$

$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }\quad$

$m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }$

$\cfrac { 1 }{ 2 } m{ \left( \cfrac { v }{ { t }_{ 1 } } \right) }^{ 2 }{ t }^{ 2 }$

$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 }^{ 2 } } { t }^{ 2 }$

$1$

$4/3$

$16/9$

$200J-500J$

$2\times 10^5J-3\times 10^5J$

$20000J-50000J$

$2000J-5000J$