Force and Motion 2

Question 1

What is the two types of friction and their differences?

Answer 1

f_static = µ_staticF_N & f_kinetic = µ_kineticF_N

Question 2

Answer 2

(a) 0 N
(b) 5 N
(c) No
(d) Yes
(e) 8 N

Question 3

Answer 3

(a) Both vectors are antiparallel
(b) Both vectors are parallel
(c) Velocity only changes by direction, magnitude stays constant. Thus the magnitude of acceleration is also constant everywhere.
(d) Considering mg, (a) & (b): The magnitude of F_N is greater at the lowest point.

Question 4

A ________ is anything that can flow—-generally either a gas or a liquid. When there is a relatice velocity between a fluid and a body, the body experiences a _________________ D that opposes the relative motionand points in the direction in which the fluid flows relative to the body.

Answer 4

fluid, Drag Force.

Question 5

Under the assumptions that we treat air as a fluid, the body is blunt, and the relative motion is fast enough so that the air becomes turbulent behind the body: what would be the equation for D.

Answer 5

D = ½CρAv², where A is the effective crossectional area of the body and ρ is the fluid density.

Question 6

Derive terminal speed.

Answer 6

Under acceleration the body has D - F_G = ma,

Under terminal speed a →0 (constant velocity).

Thus v_T = (2F_G / CρA)^½.

Question 7

If a particle moves in a circle or a circular arc of radius R at constant speed v, the particle is said to be in what type of motion?

What would be the magnitude of it’s acceleration?

Answer 7

Uniform circular motion,

a = v² / r

Question 8

The acceleration that is due to a net centripetal force on a particle is given by what equation?

Answer 8

F_net = mv² / r, where m is the particle’s mass. The vector quantities a and F are directed toward the center of curvature of the particle’s path.

Question 9

A centripetal force accelerates a body by changing the direction of the body’s __________ without changing the body’s _______.

Answer 9

velocity, speed.

Question 10

Answer 10

Forces on block B:

T_B = μm_Bg

Forces on knot:

T_B = Tcosθ & Tsinθ = T_A

Forces on block A:

T_A = m_Ag

Now solve for m_A.

Question 11

Answer 11

A) They all move at the same acceleration,

F_applied - μm_totalg = m_Ta.

Solving for a then using it in

F₃₂ = μm₃g + m₃a.

B) After some algebra,

F₃₂ = (m₃/m_total)F which is independent of friction so the answer would be the same as part A.

Question 12

Answer 12

Let us assume that +a would be in the upward direction of the incline.

Solvng for friction while everything is at rest one will see that friction is a negative value which implies that our original assumption of friction is incorrect. Nonetheless, one can calculate and see f_s,max>f_calculated ⇒ a = 0.

B) a = (W_B - W_Asinθ - μW_Acosθ) / (m_B + m_A)

C) (W_B - W_Asinθ + μW_Acosθ) / (m_B + m_A)

Question 13

Answer 13

f_s,max =59N,

x-axis: -f = m_sa_s, f - F = m_ba_b

if the block does not slide then a_s = a_b ⇒ f = 80N. Thus, it slides since 59<80 and a_s does not equal a_b.

Using the y components of forces one can solve the x component of forces for corresponding acceleration.

Question 14

Answer 14

Question 15

Answer 15