Show Me The Physics
Newton's 2nd Law
AP Physics 1 - Dynamics Module
AP Physics 1 - Dynamics Module
A) 3 Laws of Motion - 1st Law - Law of Inertia Video Instruction
If $\Sigma F_{net} = 0$, then $a = $
Because
$\Sigma F_{net} = 0$ means that the object is at rest or moving with
Ex) Free Body Diagram - Block being pulled to the right at a constant velocity. Draw all the forces:
Claim $a = 0 \text{ m/s}^2$. Forces are balanced.
Evidence Constant velocity implies $\Delta v = 0$, therefore $a = \frac{\Delta v}{t} = 0 \text{ m/s}^2$. By Newton's 2nd Law: $\Sigma F = ma = m(0 \text{ m/s}^2) = 0 \text{ N}$.
Reasoning According to the Law of Inertia, an object in motion stays in motion at a constant velocity unless acted upon by an unbalanced force. Since velocity is constant, the net force must be zero.
A net force is not required
A 5.0 kg box is moved across a floor at constant velocity with a horizontal F of 12 N. What is the force of friction on the box?
Claim The force of friction is $12 \text{ N}$ directed to the left.
Evidence $\Sigma F_x = F_{applied} - F_f = ma$. At constant velocity, $a = 0 \text{ m/s}^2$.
$12 \text{ N} - F_f = (5.0 \text{ kg})(0 \text{ m/s}^2) \implies F_f = 12 \text{ N}$.
Reasoning Because the box is in dynamic equilibrium, the net force must be zero. The magnitude of friction must match the applied force exactly.
Unbalanced forces cause objects to
$F_{net}$ = Resultant
Ex) Mass M below is lifted up with an acceleration $a$. Find the tension in the string $T$.
$\Sigma F_{net} = ma$ (up )
Claim Tension $T = M(g + a)$.
Evidence $\Sigma F_y = T - F_g = Ma$.
$T - M(10 \text{ m/s}^2) = M(a) \implies T = M(10 \text{ m/s}^2 + a \text{ m/s}^2)$.
Reasoning To accelerate upward, the upward tension must overcome weight AND provide the extra force for acceleration.
If the String is Accelerated downward?
Claim Tension $T = M(g - a)$.
Evidence $\Sigma F_y = F_g - T = Ma$.
$M(10 \text{ m/s}^2) - T = Ma \implies T = M(10 \text{ m/s}^2 - a \text{ m/s}^2)$.
Reasoning Gravity is the dominant force. Tension is reduced by the force used for downward acceleration.
Video Instruction p. 3 - 5
Ex 1) Two masses are connected by a string and remain at rest. Find tension force on top and bottom rope.
$\Sigma F = $
Claim $T_{bottom} = m_B (10 \text{ m/s}^2)$ and $T_{top} = (m_A + m_B)(10 \text{ m/s}^2)$.
Evidence Isolate $m_B$: $T_{bot} - m_B(10) = 0$. Isolate system: $T_{top} - (m_A+m_B)(10) = 0$.
Reasoning Static systems require tension to equal the weight of everything hanging below.
Ex 2) Find the tension of the top and bottom string if objects accelerated upward.
Claim $T_{bot} = m_B(10+a)$; $T_{top} = (m_A+m_B)(10+a)$.
Reasoning Both segments provide weight support plus acceleration force for the masses trailing behind them.
Which cart is slowing down as it travels to the right?
Claim Cart C is slowing down.
Reasoning Deceleration occurs when the net force (and thus acceleration) is opposite to the velocity vector.
Resolving Weight force on Ramp
Label $F_g \sin \theta$ and $F_g \cos \theta$ (Parallel & perpendicular components)
Claim $F_{||} = m(10) \sin \theta$; $F_{\perp} = m(10) \cos \theta$.
Reasoning $F_{||}$ is opposite the angle in the force triangle (sine), while $F_{\perp}$ is adjacent (cosine).
Find (f) friction if the box is stationary. CAR at rest, Find $\mu_s$.
Claim $\mu_s = \tan \theta$.
Evidence $m(10) \sin \theta = \mu_s (m(10) \cos \theta) \implies \mu_s = \frac{\sin \theta}{\cos \theta}$.
The box rests on a ramp. What is T in terms of M, $\theta$?
Claim $T = M(10 \text{ m/s}^2) \sin \theta$.
Determine the acceleration of the blocks after release.
Force of the tension of the rope on mass m, M?
Massless Pulley - Rule - One Rope,
Claim $a = \frac{(M - m) \cdot 10}{M + m}$. Rule: One Tension.
Reasoning In a massless pulley, the tension is uniform throughout the single continuous rope.
Ex 1) Find acceleration for this system (frictionless table)
Two bodies move
$F_{net} = $
Ex 2) Acceleration if smaller mass was hanging?
Isolate m (Ex 3 Tension):
Ex 4) Find a, if Box on table has $\mu_k$
Claim Ex 1: $a = \frac{m(10)}{M+m}$. Ex 4: $a = \frac{m(10) - \mu_k M(10)}{M+m}$.
Reasoning Frictionless: $F_{net} = m_{hang}(10)$. With friction: $F_{net} = m_{hang}(10) - f_k$.
Masses move
a) When does $a = 0$?
b) Under what circumstances will the large box accelerate down the ramp?
Claim $a=0$ if $m_{hang} = M_{ramp} \sin \theta$.
$F_{net} = $
$m = $
Claim $a = (10 \text{ m/s}^2) \sin \theta$.
Reasoning Frictionless acceleration on an incline is purely gravitational and independent of mass.
1) A 10.-kg cement block, 40. N push, $a = 1.0 \text{ m/s}^2$. What is $\mu$?
Claim $\mu = 0.30$.
Evidence $40 \text{ N} - f = (10 \text{ kg})(1.0 \text{ m/s}^2) \implies f = 30 \text{ N}$. $F_N = (10 \text{ kg})(10 \text{ m/s}^2) = 100 \text{ N}$. $\mu = 30 / 100 = 0.30$.
2) A 20.-kg box, $\mu = 0.50$, $a = 6.0 \text{ m/s}^2$. What is the push?
Claim $F_{push} = 220 \text{ N}$.
Evidence $F_f = (0.5)(20 \text{ kg})(10 \text{ m/s}^2) = 100 \text{ N}$. $F_{net} = (20 \text{ kg})(6.0 \text{ m/s}^2) = 120 \text{ N}$. $F_{push} = 120 + 100 = 220 \text{ N}$.
3) A 40.0-kg box, $T = 200\text{N}$ at 36.87°, $a = 2.50\text{m/s}^2$. What is $\mu$?
Claim $\mu = 0.21$.
Evidence $T_x = 160 \text{ N}, T_y = 120 \text{ N}$. $F_N = (40 \times 10) - 120 = 280 \text{ N}$. $f = 160 - (40 \times 2.5) = 60 \text{ N}$. $\mu = 60 / 280 \approx 0.214$.
1. 1050 kg car, $a = 1.20 \text{ m/s}^2$. Tension?
Claim $T = 1260 \text{ N}$.
2. Stop 1100 kg car in 8s from 90 km/hr. Force?
Claim $F_{avg} = 3437.5 \text{ N}$.
3. 1200. kg crate vertically up at 0.80 m/s². Tension?
Claim $T = 12960 \text{ N}$.
Evidence $T - (1200 \times 10) = (1200 \times 0.8) \implies T = 12000 + 960 = 12960 \text{ N}$.
4. 10.-kg bucket, $T = 63 \text{ N}$. What is $a$?
Claim $a = 3.7 \text{ m/s}^2$ down.
Evidence $(10 \times 10) - 63 = 10a \implies 37 = 10a \implies a = 3.7 \text{ m/s}^2$.
Scenario 1: Horizontal Tensions Instructional Video
Sum masses trailing behind rope segment.
Scenario 2: Vertical Tensions Instructional Video
$T = m(10+a)$.
Scenario 3: Pulley Combo Tensions Video Instruction 5 & 8
Isolate components based on direction of motion.
Scenario 4: Hanging Weight
Equilibrium logic for vertical strings.
1. $f = \mu F_n$
2. $F_n = m(10) \cos \theta$ (Ramp Perpendicular)
Kinetic Friction on ramp: $f_k = \mu m (10) \cos \theta$
3. $F_{parallel} = m(10) \sin \theta$
4. Calculate $a$ for frictionless block down ramp.
5. Does $a$ depend on mass?
Claim $a = 10 \sin \theta$; Mass has no effect.
1. 900. kg box, 25° wet slide, $\mu = 0.0500$. Find Normal, Friction, Net Force, and Accel.
Claim $a = 3.8 \text{ m/s}^2$.
Evidence $F_N = 900(10)\cos(25) = 8157 \text{ N}, F_f = 408 \text{ N}, F_{||} = 900(10)\sin(25) = 3804 \text{ N}, F_{net} = 3396 \text{ N}$.
$a = 3396 / 900 \approx 3.77 \text{ m/s}^2$.
2. Rather than taking the stairs, David likes to slide down the banister in his house. David has a mass of 20. kg, the banister has an angle of 30 degrees relative to the horizontal, and the coefficient of kinetic friction between David and the banister is 0.20. David begins his motion from rest.
Claim $a = 3.3 \text{ m/s}^2$.
Evidence $F_{||} = 20(10)\sin(30^\circ) = 100 \text{ N}, F_N = 20(10)\cos(30^\circ) = 173 \text{ N}, F_f = 0.2(173) = 34.6 \text{ N}, F_{net} = 65.4 \text{ N}$.
$a = 65.4 / 20 \approx 3.27 \text{ m/s}^2$.