Work, Energy and Power Test

Result

Work, Energy and Power Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A spot light S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light p moves along the wall at a distance 3 m. What is the velocity of the spot P when q = 45° ?

A
0.6 m/s

B
0.5 m/s

C
0.4 m/s

D
0.3 m/s

Solution

Linear Velocity of rotating object ν=ωr/sinθ
ω= angular velocity=0.1 rad/s
r is the distance of object from centre.
Angle = 45o
r = d/cosθ [where d is the distance from the spot light to wall] = 3/cos45o
Velocity = ν = ωr/sinθ
= [0.1x3/cos45degree]/sin45degree
= 0.1x3/sin45 degreecos45degree
= 0.3/[1√2 x 1 /2]
= 2x0.3
= 0.6m/s
The velocity of the spot P is 0.6 m/s
Question 2
1 / -0

Two moving particle P and Q are 10 m apart at a certain instant. The velocity of P is 8m/s making an angle 30° with the line joining P and Q and that of Q is 6m/s making an angle 30° with PQ as shown in the figure. Then angular velocity of P with respect to Q is

A
Zero

B
0.1 rad/sec

C
0.4 rad/sec

D
0.7 rad sec

Solution
Question 3
1 / -0

The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed w varies with time t as

A
2 a sin wt

B

C
2a cos wt

D

Solution

If a particle is moving with angular velocity=ω
Its angle of rotation is given by ωt
Now displacement= length of line AB
Position vector of a particle is given by
R =iacosωt + jasinωt
R_o=ai
displacement =R−R_o
=a(cosωt−1)i+asinωj
d=√[(a(cosωt-1))²+(asinω)²]
=a√(2(1-cosωt))=a√(2×2(sinωt/2)²)=2asinωt/2
Question 4
1 / -0

A particle moves along an arc of a circle of radius R. Its velocity depends on the distance covered as v = a, where a is a constant then the angle a between the vector of the total acceleration and the vector of velocity as a function of s will be

A
Tana =

B
Tana = 2s / R

C
Tana =

D
Tana =

Solution
Question 5
1 / -0

A particle A moves along a circle of radius R = 50 cm so that its radius vector r relative to the point O (Fig.) rotates with the constant angular velocity w = 0.40 rad/s. Then modulus of the velocity of the particle, and the modulus of its total acceleration will be

A
v = 0.4 m/s, a = 0.4 m/s²

B
v = 0.32 m/s, a = 0.32 m/s²

C
v = 0.32 m/s, a = 0.4 m/s²

D
v = 0.4 m/s, a = 0.32 m/s²
Question 6
1 / -0

Tangential acceleration of a particle moving in a circle of radius 1 m varies with time t as (initial velocity of particle is zero). Time after which total cceleration of particle makes and angle of 30º with radial acceleration is

A
4 sec

B
4/3 sec

C
2^2/3 sec

D
Question 7
1 / -0

A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal smooth surface. The maximum tension the string can withstand is 16 newton. The maximum speed of revolution of the stone without breaking it, will be :

A
20 ms^-1

B
16 ms^-1

C
12 ms^-1

D
14 ms^-1

Solution

T= (mv²_max)/r
16 = (16 v²_max)/144
v²_max = 12 m/s
Question 8
1 / -0

A box of mass m is released from rest at position on the frictionless curved track shown. It slides a distance d along the track in time t to reach position 2, dropping a vertical distance h. Let v and a be the instantaneous speed and instantaneous acceleration, respectively, of the box at position 2. Which of the following equations is valid for this situation?

A
h = vt

B
h = (1/2)gt²

C
mgh = (1/2)mv²

D
none

Solution

According to the Work Energy Theorem,
Change in PE = Change in KE
So, mgh = ½ mv²
Hence C
Forum id- 1775160
For complete circle v = √(5gl)
mgh = ½mv²
So, h = 2.5R