The surface of an incline is coated with an experimental substance that is intended to reduce the frictional force between a blo
ck and the
surface of the incline. A 2 kg block is placed at the top of the incline at a height of 1.8 m, as shown in the figure. After the block is released
from rest, the block slides down the incline and a motion detector at the bottom of the incline measures the block's speed as 5.8 after the
block is no longer on the incline. Which of the following claims is correct about the experimental substance?
The experimental substance reduced all the frictional force because all the gravitational potential energy of the Earth-block
system at the top of the incline was converted into the kinetic energy of the block at the bottom of the incline.
The experimental substance did not reduce all the frictional force because some of the gravitational potential energy of the
Earth-block system at the top of the incline was converted into nonmechanical energy
с
The effectiveness of the experimental substance cannot be determined because the speed of the block at the bottom of the
inline, as measured by the motion detector, indicates that the block has more energy at this location than the Earth-block
system had at the top of the incline.
D
The effectiveness of the experimental substance cannot be determined without knowing the magnitude of the frictional
force between the block and the incline before and after the experimental substance was applied to the incline.
Lo
surface of the incline. A 2 kg block is placed at the top of the incline at a height of 1.8 m, as shown in the figure. After the block is released
from rest, the block slides down the incline and a motion detector at the bottom of the incline measures the block's speed as 5.8 after the
block is no longer on the incline. Which of the following claims is correct about the experimental substance?
The experimental substance reduced all the frictional force because all the gravitational potential energy of the Earth-block
system at the top of the incline was converted into the kinetic energy of the block at the bottom of the incline.
The experimental substance did not reduce all the frictional force because some of the gravitational potential energy of the
Earth-block system at the top of the incline was converted into nonmechanical energy
с
The effectiveness of the experimental substance cannot be determined because the speed of the block at the bottom of the
inline, as measured by the motion detector, indicates that the block has more energy at this location than the Earth-block
system had at the top of the incline.
D
The effectiveness of the experimental substance cannot be determined without knowing the magnitude of the frictional
force between the block and the incline before and after the experimental substance was applied to the incline.
Lo
2 answers:
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Answer:
a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m
b) ΔU = -0.000747871 J
c) w = 47.97 rad / s
Explanation:
Given:-
- The area of the circular ring, A = 4.45 cm^2
- The current carried by circular ring, I = 13.5 Amps
- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)
- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)
- The magnetic moment final orientation, vec ( μf ) = -μ k^
- The inertia of ring, T = 6.50×10^−7 kg⋅m2
Solution:-
- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:
μ = N*I*A
Where,
N: The number of turns
I : Current in coil
A: the cross sectional area of coil
- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):
μ = 1*( 13.5 ) * ( 4.45 / 100^2 )
μ = 0.0060075 A-m^2
- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:
vec ( τi ) = vec ( μi ) x vec ( B )
= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )
= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
- Perform cross product:
- The initial torque ( τi ) is written as follows:
vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )
- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):
- The initial potential energy stored in the circular ring ( Ui ) is:
Ui = - vec ( μi ) . vec ( B )
Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]
Ui = - [(-0.000605556 + 0.00011)]
Ui = 0.000495556 J
- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :
Uf = - vec ( μf ) . vec ( B )
Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]
Uf = -0.000252315 J
- The decrease in magnetic potential energy of the ring is arithmetically determined:
ΔU = Uf - Ui
ΔU = -0.000252315 - 0.000495556
ΔU = -0.000747871 J
Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.
- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.
- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:
Ui + Eki = Uf + Ekf
Where,
Eki : The initial kinetic energy ( initially at rest ) = 0
Ekf : The final kinetic energy at second position
- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.
-ΔU = Ekf
0.5*T*w^2 = -ΔU
w^2 = -ΔU*2 / T
Where,
w: The angular speed at second position
w = √(0.000747871*2 / 6.50×10^−7)
w = 47.97 rad / s