Work Energy and Power Test

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Work Energy and Power Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Two equal masses are attached to the two ends of a spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass during the above stretching is?

A
$$\displaystyle\frac{1}{2}kx^2$$

B
$$-\displaystyle\frac{1}{2}kx^2$$

C
$$\displaystyle\frac{1}{4}kx^2$$

D
$$-\displaystyle\frac{1}{4}kx^2$$

Solution
Question 2
1 / -0

A simple pendulum with bob of mass $$m$$ and length $$x$$ is held in position at an angle $$1$$ and then angle $$2$$ with the vertical. When released from these positions, speeds with which it passes the lowest positions are $${v}_{1}$$ and $${v}_{2}$$ respectively. Then, $$\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } $$ is

A
$$\cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } $$

B
$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } } $$

C
$$\sqrt { \cfrac { 2gx(1-\cos { { \theta }_{ 1 } } ) }{ 1-\cos { { \theta }_{ 2 } } } } $$

D
$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 2gx(1-\cos { { \theta }_{ 2 } } ) } } $$

Solution

A simple pendulum with bob of mass m and length x is held in position at an angle 1 and then angle 2 with the vertical. When released from these positions, speeds with which it passes the lowest positions are $$v_1$$ and $$v_2$$ respectively.
Kinetic Energy = loss in PE
$$m v_1^2 /2 = m g L (1 - cos θ_1)$$
So,
Similarly $$m v_2^2 /2 = m g L (1 - cos θ_2)$$

Ratio $$\dfrac{v_1}{ v_2} =\sqrt{\dfrac{ (1 - cos θ_1)}{ (1 - cos θ_1)}}$$ .
Question 3
1 / -0

The work done in joules in increasing the extension of a spring of stiffness $$10\ N/cm$$ from $$4\ cm$$ to $$6\ cm$$.

A
$$1$$

B
$$10$$

C
$$50$$

D
$$100$$

Solution

Given Spring stiffness $$K= 10 N/cm= 1000 N/m$$, Initial Position $$x_1= 4 cm= 0.04 m$$, Final position $$x_2= 6 cm= 0.06 m$$
Work done by spring $$= \dfrac{1}{2}K(x_2^2-x_1^2)$$
$$\Rightarrow \text{Work done}= \dfrac{1}{2}\times 1000\times (0.06^2- 0.04^2)= 1.0 J$$
Question 4
1 / -0

A block of mass $$M$$ at the end of the string is whirled round a vertical circle of radius $$R$$. The critical speed of the block at the top of the swing is

A
$${ \left( R/g \right) }^{ 1/2 }$$

B
$$g/R$$

C
$$M/Rg$$

D
$${ \left( Rg \right) }^{ 1/2 }$$

Solution

Using conservation of energy
Total mechanical energy at lowest point = Total mechanical energy at top
$$\dfrac{1}{2}m{v_{lowest}^2} = \dfrac{1}{2}m{v_{High}^2} + 2mgR$$
we know that $$ {v_{lowest}} = \sqrt{5Rg}$$
$$\therefore \dfrac{1}{2}(5Rg) = \dfrac{1}{2}{v_{high}^2} + 2gR$$
$${v_{high}} = \sqrt{Rg}$$
Question 5
1 / -0

The mass of bucket full of water is 15 kg . It is being pulled up from a 15 m deep well. Due to a hole in the bucket 6 kg water flows out of the bucket at a uniform rate. The work done in drawing the bucket out of the well will be $$(g=10 m/s^2)-$$

A
900 J

B
1500 J

C
1800 J

D
2100 J

Solution

$$\textbf{Step 1: Water in bucket at height x} $$
Mass of bucket full of water $$ = 15\ kg$$
Water goes out of bucket $$ = 6\ kg$$
Out flow of water per meter of pull $$ = \dfrac{6kg}{15m} $$
$$ = 0.4\ kg/m$$
Water in the bucket at height x from bottom $$(m)$$ $$ = (15 - 0.4 x) kg$$
$$m = (15 - 0.4x)$$

$$\textbf{Step 2: Work done} $$
Work done in pulling the bucket by small distance $$(dx)$$
$$dW = mg\ dx $$
$$dW = (15 - 0.4 x)g\ dx $$

$$\textbf{Step 3: Integration}$$
For total work done integrate from $$x = 0$$ to $$x = 15\ m$$
$$W = \int^{15}_{0} (15 - 0.4 x) g\ x $$

$$\Rightarrow W = g \left [15 x - 0.4 \dfrac{x^2}{2} \right ]^{15}_{0} $$

$$\Rightarrow W = 10 (225 - 45) \quad$$
$$\Rightarrow W = 1800\ J$$

Option C is correct.

$$\textbf{Alternate Solution:} $$

Here work done by gravity is linear function of height $$(x)$$
So we can find average water leak in this case-
Average weight leak $$ = 3$$ g Kg - wt
So remaining $$ = (15 - 3)\ g = 12$$ g Kg - wt

Work done $$ = F \times d $$ $$ = 12\ g \times 15$$ $$ = 120 \times 15$$

$$W = 1800\ J$$

$$\textbf{Note:} $$
This method is only applicable for the linear function of height.
If work done by gravity is a non-linear function of $$x$$ then, don't use this approach
e.g. If $$W = mgx^3$$
Here we can not apply this method
Question 6
1 / -0

If the force constant of a wire is $$K$$, the work done in increasing the length of the wire by/ is

A
$$Kl/ 2$$

B
$$Kl$$

C
$$KI^{2}/2$$

D
$$Kl^{2}$$

Solution

Let initial length of the wire be L
Then final length of the wire will be L+l
Initial distance of COG of wire from the bottom $$=\dfrac{L}{2} $$

Final distance of COG of wire from the bottom $$=\dfrac{L+l}{2} $$

Displacement of the COG of the wire $$s=\dfrac{L+l}{2} $$ $$-\dfrac{L}{2} $$ $$=\dfrac{l}{2} $$

Work done $$=Ks=K\dfrac{l} {2}$$

Hence correct answer is option $$A $$
Question 7
1 / -0

A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity $$\sqrt{3gr}$$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

A
4 : 3

B
5 : 4

C
1 : 4

D
3 : 2

E
1 : 2

Solution

Tension at highest point
$$T_H=\dfrac{mv^2}{r}-mg=3mg-mg=2mg$$
and tension at lowest point
$$T_L=\dfrac{mv^2}{r}+mg$$
Here, $$V^2_L=3gr+2g.2r=7gr$$
So, $$T_L=7mg+mg=8mg$$
Hence, $$\dfrac{T_H}{T_L}=\dfrac{2mg}{8mg}=\dfrac{1}{4}$$
Question 8
1 / -0

A pendulum string of length $$l$$ is moved upto a horizontal position and released as shown in figure. If the mass of pendulum is $$m$$, then what is the minimum strength of the string that can withstand the tension as the pendulum passes through the position of equilibrium?
(neglect mass of the string and air resistance)

A
$$mg$$

B
$$2mg$$

C
$$3mg$$

D
$$5mg$$

Solution

We have, $$\quad T-mg=\cfrac { m{ v }^{ 2 } }{ I } $$
$$\quad { v }^{ 2 }=2gI\Rightarrow T=mg+\cfrac { m2gI }{ I } =3mg$$
Question 9
1 / -0

The work done by external agent in stretching a spring of force constant $$k = 100\ N/cm$$ from deformation $$x_{1} = 10$$ to deformation $$x_{2} = 20\ cm$$.

A
$$-150\ J$$

B
$$50\ J$$

C
$$150\ J$$

D
None of these

Solution

Work done by external agent =change in potential energy of spring
$$=\dfrac{1}{2}kx_2^2-\dfrac{1}{2}kx_1^2$$
$$=\dfrac{1}{2} (10000 N/m)(0.2)^2-\dfrac{1}{2}(10000)(0.1)^2$$
$$=5000(0.04-0.01)=150 J$$
Question 10
1 / -0

A particle is projected so as to just move along a vertical circle of radius $$r$$ with the help of the massless string. The ratio of the tension in the string when the particle is at the lowest and highest point on the circle is?

A
$$1$$

B
Finite but large

C
Zero

D
Infinite

Solution

$$T-mg\cos\theta=\cfrac{mV^2}{r}$$
At lowest point $$A$$:
$$T_A=\cfrac{mV_A^2}{r}+mg\cos (0^o)$$
$$\implies T_A=\cfrac{mV_A^2}{r}+mg$$
At highest point $$B$$:
$$T_B=0$$
$$\therefore \cfrac{T_A}{T_B}=\cfrac{({mV_A^2/r})+mg}{0}$$
$$\implies \cfrac{T_A}{T_B}=\infty$$

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 39

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Work Energy and Power Test - 39

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SHARING IS CARING

Question 1

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

$$-\displaystyle\frac{1}{2}kx^2$$

$$\displaystyle\frac{1}{4}kx^2$$

$$-\displaystyle\frac{1}{4}kx^2$$

Solution

Question 2

$$\cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } $$

$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } } $$

$$\sqrt { \cfrac { 2gx(1-\cos { { \theta }_{ 1 } } ) }{ 1-\cos { { \theta }_{ 2 } } } } $$

$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 2gx(1-\cos { { \theta }_{ 2 } } ) } } $$

Solution

Question 3

$$1$$

$$10$$

$$50$$

$$100$$

Solution

Question 4

$${ \left( R/g \right) }^{ 1/2 }$$

$$g/R$$

$$M/Rg$$

$${ \left( Rg \right) }^{ 1/2 }$$

Solution

Question 5

900 J

1500 J

1800 J

2100 J

Solution

Question 6

$$Kl/ 2$$

$$Kl$$

$$KI^{2}/2$$

$$Kl^{2}$$

Solution

Question 7

4 : 3

5 : 4

1 : 4

3 : 2

1 : 2

Solution

Question 8

$$mg$$

$$2mg$$

$$3mg$$

$$5mg$$

Solution

Question 9

$$-150\ J$$

$$50\ J$$

$$150\ J$$

None of these

Solution

Question 10

$$1$$

Finite but large

Zero

Infinite

Solution

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