1. Clicker Question
You are traveling on an elevator up the Sears tower. As you near the top floor and are slowing down, your acceleration A) is upward B) is downward C) is zero V a mg N

2. Checkpoint
A box of mass m is hung by a spring from the ceiling of an elevator. When the elevator is at rest the length of the spring is L = 1 m. If the elevator accelerates upward the length of the spring will be: A) L = 1 m B) L < 1 m C) L > 1 m a L m
“The box is accelerating upward, so the spring force must be greater than gravity. Therefore, L must be larger than the normal length..”

3. Accelerating reference frames can confuse…
“There is a pseudo force that pulls the block down, increasing the length of the spring” “The force from the moving elevator will make the box be pushed down, elongating the spring” “accelerating upwards would increase the force downwards and thus increasing the length of the spring” a L m
with 𝑎>0
𝐹 𝑛𝑒𝑡 =𝑚 𝑎 =𝑚𝑎 𝑦
𝐹 𝑛𝑒𝑡 = 𝐹 𝑠𝑝𝑟 + 𝐹 𝑔𝑟𝑎𝑣 = 𝑘∆𝑦−𝑚𝑔 𝑦 =𝑚𝑎 𝑦
∆𝑦 and the length is bigger for larger 𝑎
These arguments may feel good, but all you really need to remember is this: acceleration is due to a net force

5. Clicker ? [continued from Tues]
A cart with mass m2 is connected to a mass m1 using a string that passes over a frictionless pulley, as shown below. The cart is held motionless. m2
The tension in the string is
A) m1g
B) m2g
C) 0 m1 5

6. Clicker ? [continued from Tues]
A cart with mass m2 is connected to a mass m1 using a string that passes over a frictionless pulley, as shown below. Initially, the cart is held motionless, but is then released and starts to accelerate. a m2
After the cart is released, the tension in the string is
A) = m1g
B) > m1g
C) < m1g m1
m1g T a m1
acceleration is due to a net force 6

7. Checkpoint m2 m2 Case 1 g Case 2 g m1 m1
A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table. m2 m2
Case 1
(No Friction) g
Case 2
(With Friction) g m1 m1
In which case is the tension in the string biggest?
A) Case B) Case C) Same
This is a difficult question, so let’s step through it 7

8. Step 1 m2 m2
Case 1
(No Friction)
Case 2
(With Friction) m1 m1
In which case is the acceleration of the blocks biggest? A) Case 1 B) Case 2 C) Same 8

9. Step 2 m2 m2
Case 1
(No Friction)
Case 2
(With Friction) a1 m1 a2 m1
In which case is the tension in the string biggest? A) Case 1 B) Case 2 C) Same 9

10. Lets work it out
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? m2 m1 10

11. Let’s work it out
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? T1 m2 T2
A helpful aside: g
What is the relationship between the magnitude of the tension of the string at block 2 and the magnitude of the tension in the string at block 1?
A) T1 > T2 B) T1 = T2 C) T1 < T2 m1 11

12. m2 m1
1) FBD N T m2
f= mk N T m1
m2g
m1g 12

13. ACT
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? m2 g
A helpful aside: m1
What is the relationship between the magnitudes of the acceleration of the two blocks?
A) a1 = a2 B) a1 < a2 C) a1 > a2 13

14. 1) FBD 2) SF=ma N T m2 f= mk N T m1 m2g Block 2 m1g Block 1
y: N– m2g=0
N= m2g
y: T– m1g = – m1a
x: T – mk N = m2a x y
T – mk m2g = m2a 14

15. T is smaller when a is bigger
1) FBD
2) SF=ma N T m2 f T m1
m2g
m1g
T – mk m2g = m2a
Solve 2 equations
for 2 unknowns:
T– m1g = – m1a
T = m1g – m1a
a =
m1g – mk m2g
m1 + m2 x y
T is smaller when a is bigger 15