Work Energy and Power Test

Result

Work Energy and Power Test - 49

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

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half the length, the line $$OA$$ will:-

A
shift towards $$F-$$axis

B
shift towards $$X-$$axis

C
remains as it is

D
become double in length

Solution

It is known that $$ k \propto \dfrac {1}{x} \propto \dfrac {1}{L}$$

$$ \therefore k \propto \dfrac {1}{L} $$

When the length of the spring is halved, its spring constant will become double.
Slope of the force displacement graph gives the spring constant (k) of spring.
If k becomes double then slope of the graph increases, i.e. Graph shifts towards force-axis.
Question 2
1 / -0

When two springs A and B with force constant $$ K_A and K_B $$ are stretched by same force,the respective work done on them is:

A
$$ K_B : K_A $$

B
$$ K_A : K_B $$

C
$$ K_A K_B :1 $$

D
$$ \surd K\_ 4:\surd K\_ B $$

Solution

It is given that two springs A and B are there.

Work done on first spring $${{W}_{1}}=\dfrac{1}{2}{{K}_{A}}x_{1}^{2}.........(1)$$

Work done on second spring $${{W}_{2}}=\dfrac{1}{2}{{K}_{B}}x_{2}^{2}..........(1)$$

Since, force is same $$F={{K}_{A}}{{x}_{1}}={{K}_{B}}{{x}_{2}}.........(3)$$

Taking ratios of equation (1) and (2)
$$\dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{{{K}_{A}}x_{1}^{2}}{{{K}_{B}}x_{2}^{2}}=\dfrac{F{{x}_{1}}}{F{{x}_{2}}}=\dfrac{{{K}_{B}}}{{{K}_{A}}}\,\,\,\,(from\,(1))$$
Question 3
1 / -0

If two balls each of mass 0.06 Kg moving in opposite directions with speed 4 m/sec collides and rebound with the same speed,then coefficient of restitution for the collision will be:-

A
$$ \frac { 1 }{ 4 } $$

B
$$\frac { 1 }{ 2} $$

C
1

D
0

Solution

Coefficient of restitution, $$e$$

$$ e=-\dfrac{relative\,velocity\,of\,seperation}{relative\,velocity\,of\,appproach} $$

$$ e=-\dfrac{4(-\hat{i})-4(+\hat{i})}{4\hat{i}-4(-\hat{i})}=1 $$

Hence, coefficient of restitution is $$1$$
Question 4
1 / -0

A particle moves along the x- axis from x=0 to x=5 m under the influence of a force given by $$F=7-2x+3x^2$$.Work done in the process is

A
70

B
270

C
35

D
135

Solution

$$ \displaystyle w = \int F dx $$
$$ \displaystyle =\int_{0}^{5}7-2x+3x^2 \,dx $$
$$ \displaystyle = [7x-\frac{2x^2}{2}+\frac{3x^2}{3}]_{0}^{5}$$
$$ \displaystyle = [7x-x^2+x^3]_{0}^{5}$$
$$ \displaystyle = [7\times5-5^2+5^3]-[0-0+0]$$
$$ \displaystyle = [35-25+125]-0$$
$$ = 135J$$
In this question, answer is 135 J not 70J
Question 5
1 / -0

A gun fires a shell of mass $$1.5 $$kg with velocity of $$150$$ m/s and recoils with a velocity of $$2.5$$ m/s.
Calculate the mass of the gun.

A
$$20$$ kg

B
$$30$$ kg

C
$$90$$ kg

D
$$60$$ kg

Solution

Mass of shell $$=1.5$$kg
Velocity of shell $${V_1} = 159\,m/s$$
Recoil velocity $$=2.5$$ m/s
mass of gun $$=M$$
According to the conservation of momentum,
$$m{v_1} = m{v_2}$$
$$\begin{array}{l} 1.5\times 150=M\times 2.5 \\ M=\dfrac { { 1.5\times 159 } }{ { 2.5 } } \\ M=90\, kg \end{array}$$
Question 6
1 / -0

A spring of natural length $$L$$ compressed to length $$7\ L/8$$ exerts a force $$F_0$$. The work done by the spring in restoring itself to natural length is

A
$$F_0\ L/25$$

B
$$F_0\ L/16$$

C
$$3F_0\ L/25$$

D
$$F_0\ L/8$$

Solution
Question 7
1 / -0

The $$P.E.$$ and $$K.E.$$ of a helicopter flying horizontally at a height $$400\ m$$ are in the ratio $$5:2$$. The velocity of the helicopter is:

A
$$56\ m/s$$

B
$$28\ m/s$$

C
$$14\ m/s$$

D
$$42\ m/s$$

Solution

Let mass and velocity of the helicopter be $$'m'$$ and $$'v'$$
So,
$$KE=\dfrac{1}{2}mv^2$$ and
$$PE=mgh$$, where $$[h=400\ m]$$
Given ratio,
$$\dfrac{PE}{KE}=\dfrac{5}{2}$$
$$\Rightarrow \dfrac{mgh}{\dfrac{1}{2}mv^2}=\dfrac{5}{2}$$
$$\Rightarrow v^2=\dfrac{4}{5} gh$$.
$$\Rightarrow v=\sqrt{\dfrac{4}{5}\times 10\times 400}$$
$$\Rightarrow v=40\sqrt{2}\simeq 56.7\ m/s$$.
Question 8
1 / -0

At the instant t= 0 a force F=kt( k is a constant) acts on a small body of mass m resting on a smooth horizontal surface.The time,when body leaves the surface is:

A
$$mg\ k \ sin \alpha $$

B
$$\dfrac{k }{mg \ sin \alpha}$$

C
$$\dfrac{mg \ sin \alpha}{k }$$

D
$$\dfrac{mg}{k \ sin \alpha}$$

Solution
Question 9
1 / -0

Identify the wrong statement

A
A body can have momentum without energy

B
A body can have energy without momentum

C
the momentum is conserved in an elastic collision

D
Kinetic energy is not conserved in an inelastic collision

Solution
Question 10
1 / -0

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from set, the work done by the force during the first 1 sec. will be :

A
22 J

B
9 J

C
18 J

D
4.5 J

Solution

Given,
$$F=6t$$
$$m=1kg$$
Force, $$F=ma=6t$$
$$1\times \dfrac{dv}{dt}=6t$$
$$dv=6tdt$$
Integrating both side,
$$v=\int dv=\int _0^1 6tdt$$
$$v=6[\dfrac{t^2}{2}]_0^1$$
$$v=3m/s$$
Initial velocity, $$u=0m/s$$
From work energy theorem,
$$W=K_f-K_i$$
$$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$$
$$W=\dfrac{1}{2}\times 1\times 3\times 3-\dfrac{1}{2}\times 1\times 0\times 0$$
$$W=4.5-0=4.5J$$
The correct option is D.