Mechanics I

Question 1

no net force means that there is no ____ on an object

Answer 1

acceleration

Question 2

the object’s state of motion is aka it’s ____

Answer 2

velocity

Question 3

Newton’s first law = law of inertia. What is inertia?

Answer 3

inertia refers to an object’s natural resistance to change in it’s motion/velocity (an object will stay at a constant velocity unless acted upon by a net force)

Question 4

the mass of an object is the quantitative measure of ____

Answer 4

inertia

Question 5

Object 1 = 1kg. Object 2 = 100kg. Object 2 has _____ times more the inertia than object 1

Answer 5

100

Hence, it’s 100x more difficult to cause a change in object 2’s motion/velocity compared to object 1’s motion/velocity

Question 6

What is the equation for Newton’s second law?

Answer 6

F_net = ma

Question 7

When the net force is zero, the acceleration of that object is also zero? Does this mean that the object’s velocity is zero?

Answer 7

No. Velocity doesn’t have to be zero. Zero acceleration just means that the object’s velocity is not changing, rather just staying constant. It can be constantly zero or constantly a given numerical value

Question 8

F_1-on-2 = ma

Which mass is used to configure this force. Is it the mass of object 1 or 2?

Answer 8

The mass of object 2 is used to find the force exerted by object 1 onto object 2.

Question 9

Do the forces in an action-reaction pair always cancel each other?

Answer 9

No. F_1-on-2 = - F_2-on-1.

These two forces do not add up to zero because these 2 forces are not added together. For instance, the net forces applied to Object 2 include F_1-on-2, but does not include F_2-on-1. F_2-on-1 is a force applied by object 2, not on object 2; therefore it doesn’t cancel.

Question 10

Two crates are moving along a frictionless, horizontal surface. The first crate, of mass M = 100kg, is being pushed by a force of 300N. The first crate is in contact with a second crate, of mass m = 50kg.

1) What’s the acceleration of the crates?
2) What is the net force on M (F_{net on M})?

Answer 10

1) F_net = m_neta ⇒ a = F/m

a = 300N/(100kg + 50kg) = 2m/s²

2) F_{net on M} = F_crate + F_m

F_crate = 300N F_m = 50kg(2m/s²) = -100N

F_{net on M} = 300N + (-100N) = 200N

Question 11

Two crates are moving along a frictionless, horizontal surface. The first crate, of mass M = 100kg, is being pushed by a force of 300N. The first crate is in contact with a second crate, of mass m = 50kg.

A) What’s the force exerted by the larger crate on the smaller one?

B) What’s the force exerted by the smaller crate on the larger one?

Answer 11

A) F_1-on-2 = m₂a

F_1-on-2 = 50kg(2m/s²) = 100N

B) -100N

Question 12

The human body can only withstand a vertical g-force of about 5g before the body has difficulty pumping blood out of the feet and into the brain. Approximately how much upward force could be applied to a 60kg person at sea level before that person risked fainting? (The phrase g-force is a misnomer because it actually refers to acceleration: the real force involved is the normal force from the surface of contact. A person in free fall experiences “zero gees.”)

Answer 12

At free fall, a person experinces, zero gees because there is no normal force during free fall. Normal force only occurs when the person is standing on a surface. On the flat surface of the earth, the person’s F_N corresponds with 1g. F_N = 60 x 10 = 600N. Therefore, 1g = 600N. Since the person experiences 1g already just standing on the flat surface, the surface would have to push back with an additional 4g for the person to experience 5g-force. 600 x 4 = 2400N

Question 13

Equation for Force of kinetic friction

Answer 13

F_f,k = µ_kF_N

Question 14

Equation for maximum force of static friction

Answer 14

F_f,s = µ_sF_N

Question 15

You push a 50kg block of wood across a flat concrete driveway, exerting a constant force of 300N. Coefficient of static friction is 0.5, what will be the acceleration of the block?

Answer 15

F_N = 50kg(10m/s²) = 500N

F_f,s = 0.5(500N) = 250N

F_net = 300N - 250N = 50N

a = F_net/m = 50N/50kg = 1m/s²

Question 16

You pull a 50kg block with an attached rope across a flat concrete driveway, The tension in the rope has a constant force of 300N at an angle 30º to the horizontal. Coefficient of static friction is 0.5. What will the block’s acceleration be?

Answer 16

F_x = F_Tcos30 = 300N(0.85) = 255N

F_N = F_g - F_y = 50kg(10m/s²) - 300Nsin30 = 500 - 150 = 350N

F_f,s = 0.5(350N) = 175N

F_net = 255 - 175 = 80N

a = 80N/50kg = 8/5 = 1.6m/s²

Question 17

the force d/t gravity acting parallel to the inclined plane = mg____theta

Answer 17

sine

F_g = mg sin(theta)

Question 18

the force d/t gravity acting perpendicular to the inclined plane = mg___theta

Answer 18

cosine

F_N = mgcos(theta)

Question 19

A block of mass m = 4kg is placed at the top of a frictionless inclined ramp angle 30 and has a length of 10m. What is the block’s acceleration down the ramp?

Answer 19

1) a = F_net/m F_g = mgsin(theta)

In this problem, only F_g is part of the net force since the plane is frictionless (so no need to calculate/use F_N)

a = mgsin30/m = gsin30 = 10(0.5) = 5m/s²

*also note that the mass of the object is technically irrelevant because all forces were directionally proportional to mass. mass cancelled out. this is common in problems in which the forces on an object are all functions of gravity.

Question 20

A block of mass m = 4kg is placed at the top of a frictionless inclined ramp angle 30 and has a length of 10m. The block’s accerleration down the ramp is 5m/s². How long will it take for the block to slide to the bottom?

Answer 20

Use #3 of Big Five: d = v_ot + 1/2at²

v_o = 0 ⇒ d = 1/2at²

10 = 1/2 (10)t²

10/(5/2) = 2(10)/5 = 20/5 = 4

t² = 4 ⇒ t = 2s