| Name: _____________________ | Class: Physics 214 |
| SSN/ID: _____________________ | Section & Group: ____________ |
Objective
This experiment will determine the force of kinetic friction using
Newton's 2nd Law of Motion by measuring the acceleration of
an object moving on a horizontal force. The force pulling the car is
provided by the force of gravity on small masses attached to it by a
string. When these masses are small compared to the mass of the car,
friction plays a role in the experiment. The idea is that when the
mass is released, it falls to the floor (due to the force of gravity)
and accelerates the car.
Equipment
Procedure
vb - va
Acceleration a = ---------- where Δtab = tb (final) - ta (final)
Δtab
vavg = ½ (va + vb)
and vavg = Dab / Δtab , where Dab = xb (final) - xa (final)
Fnet = mnetanet = Fpulling - Ffriction
Fnet = mnetanet = m2g - Ffr
rearranging → Ffr = m2g - mnetanet
But for the total system, Newton's 2nd Law says:
Fnet = [total_mass] x [acceleration] = (M1 +
m2) a
Questions
Notes
| # | Physical Values | Car Mass = 364-gm | Car Mass = 410-gm | ||||
| 20-gm mass | 50-gm mass | 70-gm mass | 20-gm mass | 50-gm mass | 70-gm mass | ||
| 1 | xa (initial) [m] | 0.200m | 0.200m | 0.200m | 0.200m | 0.200m | 0.200m |
| 2 | ta (initial) [s] | ||||||
| 3 | xa (final) [m] | 0.250m | 0.250m | 0.250m | 0.250m | 0.250m | 0.250m |
| 4 | ta (final) [s] | ||||||
| 5 | xb (initial) [m] | 0.600m | 0.600m | 0.600m | 0.600m | 0.600m | 0.600m |
| 6 | tb (initial) [s] | ||||||
| 7 | xb (final) [m] | 0.650m | 0.650m | 0.650m | 0.650m | 0.650m | 0.650m |
| 8 | tb (final) [s] | ||||||
| 9 | Δta = ta (final) - ta (initial) [s] | ||||||
| 10 | va = 0.05m / Δtasec [m/s] | ||||||
| 11 | Δtb = tb (final) - tb (initial) [s] | ||||||
| 12 | vb = 0.05m / Δtbsec [m/s] | ||||||
| 13 | Δtab = tb (final) - ta (final) [s] | ||||||
| 14 | a = (vb - va)/Δtab [m/s2] | ||||||
| 15 | Ffr = m2g - (m1 + m2)a [N] | ||||||
Graph of Ffr vs. m2