Work Energy and Power Test

Result

Work Energy and Power Test - 48

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 body covers a distance of $$4m$$ under the action of force $$F=\left( 17-4x \right) N$$ where x is the metres. The work done by the force is

A
$$32J$$

B
$$20J$$

C
$$36J$$

D
None

Solution

Given,
$$F=(17-4x)N$$
Work done by the force, $$dW=F.dx$$
$$W=\int_0^4 (17-4x)dx=17x|_0^4-\dfrac{4x^2}{2}|_0^4$$
$$W=17(4-0)-2(16-0)$$
$$W=68-32=36J$$
The correct option is C.
Question 2
1 / -0

A force acts on a $$3$$g particle in such a way that the position of the particle as a function of time is given by $$x= 3t -4t^2+t^3$$, where $$x$$ is in meters and $$t$$ is in seconds. The work done during the first $$4$$ second is :

A
$$2.88$$ J

B
$$450$$ mJ

C
$$490$$ mJ

D
$$530$$ mJ

Solution

We have,
mass, $$m=3\ g=0.003\ kg$$
$$x=3t-4t^2+t^3$$

Now,
$$v=\dfrac{dx}{dt}=3-8t+3t^2\Rightarrow dx=(3-8t+3t^2)dt$$

$$\Rightarrow a=\dfrac{dv}{dt}=0-8+6t$$

Now,
$$dw=Fdx$$

$$\Rightarrow dw=(ma)dx$$

$$\Rightarrow dw=(0.003)(-8+6t)(3-8t+3t^2)dt$$

$$\Rightarrow dw=(0.003)(18t^3-72t^2+82t-24)dt$$

$$\Rightarrow w=(0.003)\displaystyle \int_{0}^{4}{(18t^3-72t^2+82t-24)dt}$$

$$\Rightarrow w=0.003\times 176=0.528\ J$$

$$\Rightarrow w=530\ mJ$$
Question 3
1 / -0

A particle is projected on friction less inclined plane of inclination $$90^{\circ}$$ from the horizontal, with the projection angle $$45^{\circ}$$ from the inclined plane as shown in the figure. After one collision from the plane. It reaches to its initial point of projection. Coefficient of restitution between particle and plane is

A
$$\frac{2}{3}$$

B
$$\frac{1}{3}$$

C
$$\frac{3}{25}$$

D
$$\frac{1}{\sqrt{2}}$$

Solution
Question 4
1 / -0

A $$4\ kg$$ particle moves along the $$X-$$axis. Its position $$x$$ varies with time according to $$x\left( t \right) = t + 2{t^3}$$ , where $$X$$ is in $$m$$ and $$t$$ is in seconds. Compute kinetic energy of the particle in time t

A
$$ KE=72t^2+24t+2$$

B
0

C
20t

D
40t

Solution

Kinetic energy is given by,

$$KE=\dfrac{1}{2}mv^2$$

$$v=\dfrac{dx}{dx}$$

$$=\dfrac{d}{dt}(t+2t^3)$$

$$v=1+6t^2$$

$$\Rightarrow v^2=(1+6t^2)^2=1+36t^4+12t^2$$

$$KE=\dfrac{1}{2}\times 4\times (1+36t^4+12t^2)$$

$$=2(1+36t^4+12t^2)$$

$$\therefore KE=72t^2+24t+2$$
Question 5
1 / -0

A particle of mass m strikes a wall at an a wall at angle of incidence $$60^o$$ with velocity v elastically. The change in momentum is:

A
$$mv$$

B
$$m v/2$$

C
$$-2 m v$$

D
$$Zero$$

Solution
Question 6
1 / -0

Two particles A and B equal masses are respectively tied to the centre and one end of a string, whose other end 0 is fixed. Both the particles always revolve in concentric circles of centre O. The ratio of tensions in both the parts of the string will be

A
3:2

B
2:3

C
1:2

D
1:1

Solution
Question 7
1 / -0

A stone of mass $$1\ kg$$ is tied to a string $$4\ m$$ long and is rotated at constant speed of $$40\ ms^{-1}$$ in a vertical circle. The ration of the tension at the top and the bottom is

A
11:12

B
39:41

C
41:39

D
12:11

Solution
Question 8
1 / -0

Circular flexible current loop of radius R carrying current I is placed in an inward magnetic field B. If we spin the loop with angular speed $$/omega $$, then tension in string

A
Is zero

B
is more than iBR

C
is less than iBR

D
Does not depend on rotation

Solution
Question 9
1 / -0

The disc is at rest at the top of a rough inclined plane. It rolls without slipping. At the bottom of inclined plane there is a vertical groove if radius $$R$$. In order to loop the groove, the minimum height of incline required is:

A
$$\dfrac{15 R}{4}$$

B
$$\dfrac{9 R}{4}$$

C
$$\dfrac{5 R}{2}$$

D
$$\dfrac{7 R}{5}$$

Solution

Hence, option $$(A)$$ is correct answer.
Question 10
1 / -0

A horizontal $$50\ N$$ force acts on a $$2\ kg$$ crate which is at rest on a smooth horizontal surface. At the instant the particle has gone $$2\ m$$, the rate at which the force is doing work is

A
$$2.5\ W$$

B
$$25\ W$$

C
$$100\ W$$

D
$$500\ W$$

Solution