Work Energy and Power Test

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Work Energy and Power Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

The scalar product of two vectors A and B is

A
a scalar

B
vector

C
a complex number

D
a tensor

Solution

Scalar product means dot product and dot product of 2 vectors gives a scalar , example dot product of force and displacement gives work which is scalar
Question 2
1 / -0

The scalar product of two vectors A and B in terms of the projections of the vectors on the x, y and z axis is

A
AB = A_xB_x + A_yB_y + A_zB_z

B
AB = A_xB_x + A_yB_y − A_zB_z

C
AB = A_xB_x − A_yB_y − A_zB_z

D
AB = A_xB_x − A_yB_y + A_zB_z

Solution

A = A_x i+ A_y j+ A_zk

B = B_x i+ B_y j+ B_zk

A. B = (A_x i+ A_y j+ A_zk). ( B_x i+ B_y j+ B_zk)

A. B = A_xB_x + A_yB_y+ A_zB_z
Question 3
1 / -0

The work done by the force is defined to be

A
the product of force and the magnitude of the displacement

B
the product of component of the force in the direction of the displacement and the magnitude of the displacement

C
the negative product of component of the force in the direction of the displacement and the magnitude of this displacement

D
the product of component of the force in the direction perpendicular to displacement and the magnitude of the displacement

Solution

Work done is given by

W = (Fcosθ)d

here Fcosθ is the component of applied force in direction of displacement and d is magnitude of displacement.
Question 4
1 / -0

In which case is the work done zero?

A
Force and displacement are perpendicular to each other

B
Force and displacement are in the same direction

C
Force and displacement are at an angle of 45^∘

D
Force and displacement are at an angle of 75^∘

Solution

Work done is given as

W = Fdcosθ

Here θ is the angle between F and d if both are perpendicular then θ = 90 degree so cosθ = 0 and thus work done is 0 .
Question 5
1 / -0

The S.I unit of force is

A
Joule

B
dyne

C
erg

D
Newton

Solution

In International System of Units (SI) the newton is the unit for force. It is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of 1 m/sec² in direction of the applied force. It is named after Isaac Newton in recognition of his work on classical mechanics, specifically Newton's second law of motion.

1N = 1Kgm/sec²

Dyne is a cgs unit of force. One dyne is equal to 10⁻⁵ N
Question 6
1 / -0

In a graph of F(x) vs x, the area under the curve

A
represents the momentum of F(x).

B
represents the impulse of F(x)

C
represents energy of F(x).

D
represents work done by F(x).

Solution

Work done by a variable force is given by

W = ∫ F(x)dx

above integration gives us area area under F and x.

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Work Energy and Power Test - 4

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Work Energy and Power Test - 4

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Question 1

a scalar

vector

a complex number

a tensor

Solution

Question 2

AB = AxBx + AyBy + AzBz

AB = AxBx + AyBy − AzBz

AB = AxBx − AyBy − AzBz

AB = AxBx − AyBy + AzBz

Solution

Question 3

the product of force and the magnitude of the displacement

the product of component of the force in the direction of the displacement and the magnitude of the displacement

the negative product of component of the force in the direction of the displacement and the magnitude of this displacement

the product of component of the force in the direction perpendicular to displacement and the magnitude of the displacement

Solution

Question 4

Force and displacement are perpendicular to each other

Force and displacement are in the same direction

Force and displacement are at an angle of 45∘

Force and displacement are at an angle of 75∘

Solution

Question 5

Joule

dyne

erg

Newton

Solution

Question 6

represents the momentum of F(x).

represents the impulse of F(x)

represents energy of F(x).

represents work done by F(x).

Solution

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AB = A_xB_x + A_yB_y + A_zB_z

AB = A_xB_x + A_yB_y − A_zB_z

AB = A_xB_x − A_yB_y − A_zB_z

AB = A_xB_x − A_yB_y + A_zB_z

Force and displacement are at an angle of 45^∘

Force and displacement are at an angle of 75^∘