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SVEN [57.7K]
3 years ago
15

Consider the following statements about Newton's 2nd law in general. Select all of the statements that are true. Note, there may

be more than one. Pay attention to synonyms: velocity and change in position are the same thing.
1. A net force causes velocity to change
2. If an object has a velocity, then we can conclude that there is a net force on the object
3. Accelerations are caused by the presence of a net force
Physics
1 answer:
Art [367]3 years ago
8 0

Answer:

1. True

2. False

3. True

Explanation:

Newton's 2nd law states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

\sum F = ma (1)

where

\sum F is the net force on the object

m is its mass

a is the acceleration

Furthermore, we know that acceleration is defined as the rate of change of velocity:

a = \frac{dv}{dt}

So let's now analyize the three statements:

1. A net force causes velocity to change: TRUE. Net force (means non-zero) causes a non-zero acceleration, which means that the velocity of the object must change.

2. If an object has a velocity, then we can conclude that there is a net force on the object: FALSE. The fact that the object has a velocity does not imply anything about its acceleration: in fact, if its velocity is constant, then its acceleration is zero, which would mean that the net force on the object is zero. So this statement is not necessarly true.

3. Accelerations are caused by the presence of a net force: TRUE. This is directly implied by eq.(1): the presence of the net force results in the object having a non-zero acceleration.

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A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m
Dvinal [7]

(a) 638.4 J

The work done by a force is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so \theta=0^{\circ}. Therefore, the work done by this force is

W_F=(152)(4.2)(cos 0)=638.4 J

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

W=(19.6)(9.8)=192.1 N

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}

Where \alpha = 24^{\circ} is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

R=mg cos \alpha

And substituting

m = 19.6 kg

g = 9.8 m/s^2

\alpha=24^{\circ}

We find

R=(19.6)(9.8)(cos 24)=175.5 N

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

\theta=90^{\circ}

Therefore, the work done by the normal force is

W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0

d) -194.5 J

The magnitude of the force of friction is

F_f = \mu R

where

\mu = 0.264 is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

F_f = (0.264)(175.5)=46.3 N

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

\theta=180^{\circ}

Therefore, the work done by friction is

W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

W_F=(152)(3.80)(cos 0)=577.6 J

W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J

W_R=(175.4)(3.80)(cos 90)=0

W_f =(46.3)(3.80)(cos 180)=-175.9 J

So the total work done is

W=577.6+(-296.9)+0+(-175.9)=104.8 J

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s

6 0
3 years ago
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