Mechanics I Flashcards

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1
Q

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

A

acceleration

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2
Q

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

A

velocity

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3
Q

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

A

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)

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4
Q

the mass of an object is the quantitative measure of ____

A

inertia

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5
Q

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

A

100

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

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6
Q

What is the equation for Newton’s second law?

A

Fnet = ma

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7
Q

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

A

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

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8
Q

F1-on-2 = ma

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

A

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

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9
Q

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

A

No. F1-on-2 = - F2-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 F1-on-2, but does not include F2-on-1. F2-on-1 is a force applied by object 2, not on object 2; therefore it doesn’t cancel.

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10
Q

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 (Fnet on M)?

A

1) Fnet = mneta ⇒ a = F/m

a = 300N/(100kg + 50kg) = 2m/s2

2) Fnet on M = Fcrate + Fm

Fcrate = 300N Fm = 50kg(2m/s2) = -100N

Fnet on M = 300N + (-100N) = 200N

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11
Q

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?

A

A) F1-on-2 = m2a

F1-on-2 = 50kg(2m/s2) = 100N

B) -100N

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12
Q

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.”)

A

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 FN corresponds with 1g. FN = 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

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13
Q

Equation for Force of kinetic friction

A

Ff,k = µkFN

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14
Q

Equation for maximum force of static friction

A

Ff,s = µsFN

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15
Q

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?

A

FN = 50kg(10m/s2) = 500N

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

Fnet = 300N - 250N = 50N

a = Fnet/m = 50N/50kg = 1m/s2

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16
Q

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?

A

Fx = FTcos30 = 300N(0.85) = 255N

FN = Fg - Fy = 50kg(10m/s2) - 300Nsin30 = 500 - 150 = 350N

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

Fnet = 255 - 175 = 80N

a = 80N/50kg = 8/5 = 1.6m/s2

17
Q

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

A

sine

Fg = mg sin(theta)

18
Q

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

A

cosine

FN = mgcos(theta)

19
Q

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?

A

1) a = Fnet/m Fg = mgsin(theta)

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

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.

20
Q

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?

A

Use #3 of Big Five: d = vot + 1/2at2

vo = 0 ⇒ d = 1/2at2

10 = 1/2 (10)t2

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

t2 = 4 ⇒ t = 2s

21
Q
A