1. Whiteboard Grading Readability (size, neatness, color)
Organization (flow of work)
Physics (general eqns, diagram, units)
Presentation (does presenter know what the group did on the board)
On task (during preparation time, presentation time)

3. Ft FN Ft Fg Fg 3rd Law Pair on hanging mass on cart by track by cart
by hanging mass Fg
on hanging mass
by Earth Fg
on cart
by Earth

4. Fg hanging OR Fg hanging on system by Earth on system
Dot represents BOTH cart and hanging mass

5. What if there is friction?
Fg hanging
on system
by Earth
Ffk cart
on system
by track OR
Ffk cart
on system
by track
Fg hanging
on system
by Earth

6. Mass of glider: 0.3 kg
Hanging mass: 0.05 kg
Friction = 0.1 N
Calculate the acceleration of the cart/mass system.
Calculate the tension in the string.

7. Fg hanging Ffk cart Fg hanging Ffk cart 𝐹 𝑔 = 9.8 𝑁 𝑘𝑔 0.05 kg =0.49N
on system
by Earth
Ffk cart
by track
Fg hanging
on system
by Earth
Ffk cart
by track
𝐹 𝑔 = 9.8 𝑁 𝑘𝑔 kg =0.49N
𝐹 𝑔 = −9.8 𝑁 𝑘𝑔 kg =−0.49N
𝐹 𝑓𝑘 =−0.1N
𝐹 𝑓𝑘 =0.1N
𝑎= 𝐹 𝑛𝑒𝑡 𝑚 = 0.49𝑁−0.1𝑁 0.3𝑘𝑔+0.05𝑘𝑔 = 0.39𝑁 0.35𝑘𝑔 =1.1 𝑚 𝑠 2
𝑎= −0.49𝑁+0.1𝑁 0.3𝑘𝑔+0.05𝑘𝑔 = −0.39𝑁 0.35𝑘𝑔 =−1.1 𝑚 𝑠 2

8. Now how do we calculate the tension in the string?
Why can’t we use the force diagrams we just used?

9. Ft FN Ffk Ft Fg Fg 3rd Law Pair on hanging mass on cart by track
by cart
3rd Law Pair
Ffk
on cart
by track Ft
on cart
by hanging mass Fg
on hanging mass
by Earth Fg
on cart
by Earth

10. FN Ffk Ft Fg 𝑎= 𝐹 𝑓𝑘 + 𝐹 𝑇 𝑚 𝑐𝑎𝑟𝑡 1.1 𝑚 𝑠 2 = −0.1𝑁+ 𝐹 𝑇 0.3𝑘𝑔
𝑎= 𝐹 𝑓𝑘 + 𝐹 𝑇 𝑚 𝑐𝑎𝑟𝑡 FN
on cart
by track
1.1 𝑚 𝑠 2 = −0.1𝑁+ 𝐹 𝑇 0.3𝑘𝑔
Ffk
on cart
by track Ft
on cart
by hanging mass
0.33𝑘𝑔 𝑚 𝑠 2 =−0.1𝑁+ 𝐹 𝑇
0.43𝑁= 𝐹 𝑇
Dot represents JUST the cart Fg
on cart
by Earth

11. Ft Fg 𝑎= 𝐹 𝑔 + 𝐹 𝑇 𝑚 ℎ𝑎𝑛𝑔𝑖𝑛𝑔 −1.1 𝑚 𝑠 2 = −0.49𝑁+ 𝐹 𝑇 0.05𝑘𝑔
on hanging mass
by cart
𝑎= 𝐹 𝑔 + 𝐹 𝑇 𝑚 ℎ𝑎𝑛𝑔𝑖𝑛𝑔
0.435𝑁= 𝐹 𝑇
−1.1 𝑚 𝑠 2 = −0.49𝑁+ 𝐹 𝑇 0.05𝑘𝑔
Dot represents JUST the hanging mass
−0.055𝑘𝑔 𝑚 𝑠 2 =−0.49𝑁+ 𝐹 𝑇
−0.49𝑁= 𝐹 𝑔 Fg
on hanging mass
by Earth
0.435𝑁= 𝐹 𝑇