Q1 : A mass ( m1 ) of 1.20 kg is situated at the bottom of an incline that is 30o to the horizontal. At the top of the incline, there’s a 0.500 kg disk that is kept fixed by a frictionless axle. A cord is placed over the disk and used to lift a 2.00 kg mass ( m2 ) up by a distance of 0.750 m. The free end of the cord is then attached securely to the ( m1 ) mass. The mass ( m2 ) is then released, and a coefficient of kinetic friction ( uk ) gives rise to a force of friction ( Ff ) that opposes the forward motion of the system. What speed will mass ( m2 ) have obtained when it reaches the surface below?
A: In order to answer the question, we must relate the initial and final energy states of the system. One approach to doing so is to evaluate all of the forces at work in the system:
F = ma
Fnet = F – Ff
Mass ( m2 ) has a force acting upon it in the vertical direction, but its forward progress is inhibited by the force of friction acting in the opposite direction. The force of friction, however, is not the only force that opposes the motion of ( m2 ). As we will see, a component of force due to the weight of ( m1 ) must be accounted for as well:
F – Ff – Fw = ma
If we multiply each term by the distance traveled by the system, we have a rudimentary expression for the system’s before-and-after energetics:
W = Fd
Fd – Ffd – Fwd = mad
The first term of the expression above represents the initial potential energy ( PEi ) of the system. If the the ( m2 ) mass fell unopposed to any forces, all of the energy would be converted to kinetic energy ( KEf ):
PEi = KEf
When we account for the energy barriers that must be overcome as ( m2 ) moves downward, we have the following expression:
Fd – Ffd – Fwd = ½ mv2
The expression is almost complete. What makes it incomplete is the mass term of the kinetic energy expression. In the final scenario, both masses ( m1 ) and ( m2 ) will be moving, and we must account for this. Additionally, it will be convenient to begin associating each term with their corresponding masses:
F2d – Ffd – F1wd = ½ ( m1 + m2 )vf2
d = h
m2gh – Ffh – F1wd = ½ ( m1 + m2 )vf2
The force of friction term is a product of the normal force ( N ) acting upon ( m1 ) and the coefficient of kinetic friction ( uk ):
Ff = ukN1
m2gh – ukN1h – F1wd = ½ ( m1 + m2 )vf2
If ( m1 ) were sitting on a flat surface, the normal force would be equal to its weight; however, ( m1 ) is on an incline. As a consequence, its associated weight vector must be broken down into components using trigonometry:
As we know, this is not the scenario we are dealing with. The mass will be tilted sideways by 30o in the following manner:
If we break the ( F1w ) into its component vectors, we obtain terms that can be substituted for the normal force. In addition to this, the component of ( F1w ) that opposes the motion of the system is parallel to the surface of the incline; thus, this component will be substituted for the ( F1w ) term:
N = m1g cos θ
m2gh – ( ukm1g cos θ)h – Fwd = ½ ( m1 + m2 )vf2
F1w = m1g sin θ
We now have the following expression that relates the before-and-after energetics of the system. From here on out, careful algebraic manipulation of the equation will get us where we need to be:
m2gh – ( ukm1g cos θ )h – ( m1g sin θ )h = ½ ( m1 + m2 )vf2
( gh )( m2 – ukm1 cos θ – m1 sin θ ) = ½ ( m1 + m2 )vf2
( 2gh )( m2 – ukm1 cos θ – m1 sin θ ) = ( m1 + m2 )vf2
[ ( 2gh )( m2 – ukm1 cos θ – m1 sin θ ) ] / ( m1 + m2 ) = vf2
vf = √ [ ( 2gh )( m2 – ukm1 cos θ – m1 sin θ ) ] / ( m1 + m2 )
If the block had simply fallen from some height ( h ), the final velocity would have been quantified by the ( 2gh ) term alone:
PEi = KEf
mgh = ½ mv2
gh = ½ v2
v2 = 2gh
v = √ 2gh
Let’s substitute the variables with their associated values and derive the answer we’re looking for:
vf = √ [ ( 2gh )( m2 – ukm1 cos θ – m1 sin θ ) ] / ( m1 + m2 )
vf = √ [ ( 2 )( 9.81 m/s2 )( 0.75 m )[ 2.00 kg – 1.20 kg( ukcos 30o – sin 30o ) ] / ( 1.20 kg + 2.00 kg )
vf = √ [ ( 2 )( 14.7 m2/s2 )[ 2.00 kg – 1.20 kg( 0.866uk – 0.500 ) ] / ( 3.20 kg )
uk = 0.250
vf = √ [ ( 2 )( 14.7 m2/s2 )[ 2.00 kg – 1.20 kg( 0.217 – 0.500 ) ] / ( 3.20 kg )
vf = √ [ ( 2 )( 14.7 m2/s2 )[ 2.00 kg – 1.20 kg( – 0.284 ) ] / ( 3.20 kg )
vf = √ [ ( 2 )( 14.7 m2/s2 )( 2.00 kg – 0.341 kg ) ] / ( 3.20 kg )
vf = √ [ ( 2 )( 14.7 m2/s2 )( 1.66 kg ) ] / ( 3.20 kg )
vf = √ [ ( 2 )( 24.4 kg*m2/s2 ) ] / ( 3.20 kg )
vf = √ ( 48.8 kg*m2/s2 / 3.20 kg )
vf = √ ( 15.2 m2/s2 )
vf = 3.90 m/s
Q2 : What is the tension in each section of the cord?
A:
We begin by evaluating each block individually. Here, we have a force pair, but they are not equal and opposite to one another!!! If these vectors were equal in magnitude, the mass would be stationary. For this reason, the tension ( T ) must be less than the ( m2g ) term by some factor. If ( m2 ) was in free fall, the force acting upon it would be ( m2g ), but we must remember that the entire system is accelerating with a value of ( a ), and that value must be less than ( g ):
m2g – m2a = T1
T1 = m2( g – a )
T1 = ( 2.00 kg )( 9.81 m/s2 – a )
What is the acceleration ( a ) of the system? It is the net force set equal to the product of the system’s mass and acceleration. Keep in mind that the friction force is a consequence of the normal force which acts perpendicular to the incline. Although it causes the system to lose energy, the normal force is not collinear with the force that drives the system forward. For this reason, the force of friction is not included in this derivation:
F2 – F1w = ma
F2 – F1w = ( m1 + m2 )a
a = [ ( F2 – F1w ) / ( m1 + m2 ) ]
a = ( m2g – m1g sin θ ) / ( m1 + m2 )
a = [ g( 2.00 kg – 1.20 kg sin θ ) ] / ( 3.20 kg )
a = [ g( 2.00 kg – 0.600 kg ) ] / ( 3.20 kg )
a = [ g( 1.40 kg ) ] / ( 3.20 kg )
a = g( 0.438 )
a = 4.29 m/s2
T1 = ( 2.00 kg )( 9.81 m/s2 – 4.29 m/s2 )
T1 = ( 2.00 kg )( 5.52 m/s2 )
T1 = 11.0 N
Check:
F2 = m2g
F2 = ( 2.00 kg )( 9.81 m/s2 )
F2 = 19.6 N
This value is greater than the value of T1, which is the condition for net motion of the system to occur. For this reason, the answer is reasonable:
F2 – T1 = ma
a = [ ( 19.6 N – 11.0 N ) / 2.00 kg ]
a = 4.30 m/s2
The second tension ( T2 ) value is determined in a similar fashion, but we must now use the vectors associated with mass ( m1 ) which is situated on the incline. We must also keep in mind that the tension forces are forces that keep the system’s blocks within a fixed distance from one another. This relationship is established by the strength of the cord independent of the coefficient of kinetic friction:
T2 – F1w = m1a
T2 = F1w + m1a
T2 = m1g sin θ + m1a
T2 = m1( g sin θ + a )
T2 = ( 1.20 kg )( 4.91 m/s2 + 4.29 m/s2 )
T2 = ( 1.20 kg )( 9.20 m/s2 )
T2 = 11.0 N
RENAISSANCE PHYSICS 