Download Determine the change in the speed of block A

HS Physics Wiki Quiz
Conservation of Energy
Question
Two blocks, A and B (with mass 50 kg and 100 kg,
respectively), are connected by a string, as shown here
The pulley is frictionless and of negligible mass. The
coefficient of kinetic friction between block A and the
incline is μk= 0.25.
•
Determine the change in the speed of block A
as it moves from (C) to (D), a distance of 20 m
up the incline if the system starts from rest.
Answer
12.1 m/s
Note: I used 3-4-5 triangle relationships for the sine and cosine function. You might get an answer that is
slightly different by using a calculator.
The second solution uses the calculator approach. It calculates both Kinetic Energy and Velocity.
First. Make sure you know which forces are causing this system to move.
This is about energy, so ignore the tension force in the middle. It is not adding or removing energy to the
system.
There is one non conservative force (friction) that you will need to deal with.
So we'll use the formula that allows changes in TME caused by non-conservative forces.
There are two solutions I made:
Solution A: The pretty math solution that takes a long time.
Solution B: The faster solution that you can only do if you understand what’s going on. Sometimes it is
faster to plug in numbers before you find the elegant math solution. This is probably one of those times!
Solution A
Solution B
The normal force the incline exerts on block
A is nA
=
( m g) cos37° , and the friction force is
A
=
f k µ=
µk mA g cos37° . The vertical distance block A rises is
k nA
=
∆y A
m ) sin 37°
( 20=
12 m , while the vertical displacement of block B is ∆yB =
−20 m .
We find the common final speed of the two blocks by use of
(
Wnc = KE + PEg
) − ( KE + PE )
f
g
i
= ∆KE + ∆PEg
1

This gives − ( µ k mA g cos37=
°) s  ( mA + mB ) v 2f − 0 +  mA g ( ∆y A ) + mB g ( ∆yB ) 
2

or
v 2f =
2 g  − mB ( ∆yB ) − mA ( ∆y A ) − ( µk mA cos37° ) s 
mA + mB
2( 9.80 m s2 )  − ( 100 kg ) ( −20 m ) − ( 50 kg ) ( 12 m ) − 0.25( 50 kg ) ( 20 m ) cos37°
=
150 kg
which yields v 2f = 157 m 2 s2 and a speed of 12.5 m/s. This is slightly different from the
solution A answer because of the value I used for g in this solution.
Just FYI,
The change in the kinetic energy of block A is then
∆KEA =
(
)
1
1
mA v 2f − 0=
( 50 kg ) 157 m 2 s2 = 3.9 × 103 J=
2
2
3.9 kJ