Work Energy and Power Test

Result

Work Energy and Power Test - 15

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A ball tied to a string is swung in a vertical circle. Which of the following remains constant?

A
Tension in the string.

B
Speed of the ball.

C
Centripetal force.

D
Earth's pull on the ball.

Solution

Earth's pull on the ball remains constant $$(W=mg)$$.
Tension in the string changes along the vertical circular path.
Speed of the ball changes as sum of kinetic energy and potential energy remains constant during motion.
Speed changes, hence, centripetal force changes.
Question 2
1 / -0

The K.E. of a body can be increased maximum by doubling its:

A
mass

B
weight

C
speed

D
density

Solution

$$ K.E = \dfrac{1}{2}mv^2$$

$$ K.E \propto m $$ & $$ K.E \propto v^2 $$
So doubling mass will double the kinetic energy and doubling speed will make kinetic energy $$4$$ times.

$$ \therefore $$ The K.E. of a body can be increased maximum by doubling its speed.
Question 3
1 / -0

A lighter body moving with a velocity $$v$$ collides with a heavier body at rest. Then :

A
the lighter body rebounces with twice the velocity of bigger body

B
the lighter body retraces its path with the same velocity in magnitude

C
the heavier body does not move practically

D
both (2) and (3)

Solution

In a collision system where $$m_1$$ moves with $$u_1$$ initially and $$m_2$$ is at rest with $$m_2 >>> m_1$$.
Let the final velocity of $$m_1$$ be $$v_1$$ and $$m_2$$ be $$v_2$$.
Assumption: Let the collision be elastic. Using linear momentum conservation and equation for coefficient of restitution,
$$ m_1u_1 = m_2v_2 + m_1v_1 $$
$$ v_2 - v_1 = u_1 $$

We get $$ v_1 = \dfrac {m_1 - m_2 }{m_1 + m_2} u_1 $$

$$ v_2 = \dfrac {2m_1}{m_1 + m_2} u_1 $$

Using $$m_2 >>> m_1,$$ we get $$ v_1 = -u_1 $$ and $$ v_2 = 0 $$.
Question 4
1 / -0

A certain force acting on a body of mass 2kg increase its velocity from 6m/s to 15 m/s in 2s. The work done by the force during this interval is ?

A
27J

B
3J

C
94.5J

D
189J

Solution

we know that
v=u+at
15=6+a(2)
$$a= 4.5 m/s^2$$
$$s=ut+0.5at^2= 6(2)+ 0.5(4.5)(4)= 21 m$$
$$W= mas= 2(4.5)(21)= 189 J$$
Question 5
1 / -0

Name the type of energy (kinetic energy $$K$$ or potential energy $$U$$) possessed in a compressed spring:

A
$$U$$

B
$$K$$

C
Both $$U$$ and $$K$$

D
None

Solution

When you compress a spring, it possess potential energy. The force of compression is proportional to the compression, according to Hooke's Law. Releasing the spring turns the potential energy into kinetic energy. The spring can be then used to propel some object.
Question 6
1 / -0

Energy equals of a mass of one microgram in kilo joules is

A
$$9 \times 10^{7} kJ$$

B
$$10 \times 10^3 kJ$$

C
$$8 \times 10^2 kJ$$

D
$$7 \times 10^4 kJ$$

Solution

Given
Energy of 1 microgram=?
Solution
We will use Energy mass equivalence
$$E=mc^{2}$$
$$E=10^{-6}kg\times (3 \times10^{8})^{2}m/s^{2}$$
$$E=9 \times 10^{10}J$$
Option $$A$$ is correct.
Question 7
1 / -0

A particle moves under the effect of a force $$F=Cx$$ form $$x=0$$ to $$x=x_{1}$$. The work done in the process is-

A
$$Cx_{1}^{2}$$

B
$$\displaystyle \frac{1}{2}Cx_{1}^{2}$$

C
$$Cx_{3}^{2}$$

D
zero

Solution

Similar to a spring with neutral point at x=0
$$W = \int \bar F.\bar ds = \int_0^{x_1} Cx dx = 0.5Cx_1^2$$
Question 8
1 / -0

A stone tied to a string is rotated in a vertical circle. The minimum speed with which the stone has to be rotated in order to complete the circle :

A
decreases with increasing the mass of the stone

B
is independent of the mass of the stone

C
decreases with increasing the length of the string

D
is independent of the length of the string

Solution

$$\textbf{Step 1: Energy Conservation [Ref. Fig.]}$$
$$\text{Dependence on m}$$
Let $$u$$ be the minimum speed required at $$B$$.
Apply energy conservation between point $$A$$ and $$B$$ we get
$$KE_A + PE_A = KE_B + PE_B$$
Taking point $$A$$ as zero potential level
$$\Rightarrow$$ $$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} mv^2 $$ $$....(1)$$

From equation $$1$$, we observe that mass $$m$$ is cancelled on both sides.
$$\Rightarrow u$$ is independent of $$m$$.

$$\textbf{Steo 2: Solving Equation}$$
$$\text{Dependence on R}$$
At highest point, Tension will be zero for minimum velocity
$$\Rightarrow$$ gravitational pull will provide necessary centripetal force

$$\Rightarrow mg = \dfrac{mv^2}{R}$$

$$\Rightarrow v^2 = Rg$$ .... Putting in equation $$(1)$$

We get, $$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} m (Rg)$$

$$\Rightarrow u = \sqrt{5Rg}$$

$$\Rightarrow u$$ is proportional to the square root of $$R$$.
Hence option $$B$$ is correct.
Question 9
1 / -0

Complete the following sentence:
The kinetic energy of a body is the energy by virtue of its______ .

A
force

B
motion

C
roughness

D
all the above

Solution

The kinetic energy of a body is the energy by virtue of its motion. This is the definition of kinetic energy. If it has motion then it will have velocity, hence kinetic energy exists.

$$ K.E = \dfrac{1}{2}mv^2 $$
Question 10
1 / -0

A simple pendulum, composed of a bob of mass $$m$$ connected to the end of a massless rod, executes simple harmonic motion as it swings through small angles of oscillation.The maximum angular displacement with the vertical is denoted by $${ \theta }_{ max }$$. Frictional effects and variation of acceleration due to gravity are negligible.
Find out the correct statement?

A
At $$\theta=0$$, the tangential acceleration is $$0$$

B
At $$\theta={\theta}_{max}$$, the tangential acceleration is $$0$$

C
At $$\theta=0$$, the speed is $$0$$

D
At $$\theta=0$$, the restoring force is maximized

E
At $$\theta={\theta}_{max}$$, the speed is maximized

Solution

We know the tangential acceleration in SHM is given by
$$a=-{\omega}^{2}A\sin{\theta}$$
where $$A=amplitude$$
therefore at $$\theta=0$$
$$a=-{\omega}^{2}A\sin{0^0}$$
or $$a=0$$ because $$\sin{0^0}=0$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 15

Result

Work Energy and Power Test - 15

Score

-

Rank

-

SHARING IS CARING

Question 1

Tension in the string.

Speed of the ball.

Centripetal force.

Earth's pull on the ball.

Solution

Question 2

mass

weight

speed

density

Solution

Question 3

the lighter body rebounces with twice the velocity of bigger body

the lighter body retraces its path with the same velocity in magnitude

the heavier body does not move practically

both (2) and (3)

Solution

Question 4

27J

3J

94.5J

189J

Solution

Question 5

$$U$$

$$K$$

Both $$U$$ and $$K$$

None

Solution

Question 6

$$9 \times 10^{7} kJ$$

$$10 \times 10^3 kJ$$

$$8 \times 10^2 kJ$$

$$7 \times 10^4 kJ$$

Solution

Question 7

$$Cx_{1}^{2}$$

$$\displaystyle \frac{1}{2}Cx_{1}^{2}$$

$$Cx_{3}^{2}$$

zero

Solution

Question 8

decreases with increasing the mass of the stone

is independent of the mass of the stone

decreases with increasing the length of the string

is independent of the length of the string

Solution

Question 9

force

motion

roughness

all the above

Solution

Question 10

At $$\theta=0$$, the tangential acceleration is $$0$$

At $$\theta={\theta}_{max}$$, the tangential acceleration is $$0$$

At $$\theta=0$$, the speed is $$0$$

At $$\theta=0$$, the restoring force is maximized

At $$\theta={\theta}_{max}$$, the speed is maximized

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.