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4 years ago

What Kinetic Energy does the sack have just before it strikes the floor?

1 answer:

8 0

When an object is at any height above the ground, it has gravitational potential energy. If the object falls, it loses this gravitational potential energy and this energy is converted to kinetic energy. By this logic, we can say:

Gravitational potential energy of grain sack = Kinetic energy of grain sack

GPE = mgh
        = (mg) x h
        = 98.0 x 50.0
        = 4,900 J

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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

Four spheres with positive and negative charges hang from strings.

Arada [10]

Be because it’s the answer

5 0

3 years ago

What do positive ions tend to do

notsponge [240]

Answer:

an electrically charged atom or group of atoms formed by the loss or gain of one or more electrons, as a cation (positive ion), which is created by electron loss and is attracted to the cathode in electrolysis, or as an anion (negative ion), which is created by an electron gain and is attracted to the anode.

7 0

3 years ago

When a pendulum with a period of 2.00000 s is moved to a new location from one where the acceleration due to gravity was 9.80 m/

Ivahew [28]

Answer:

0.021 $m/s^2$

Explanation:

The period of a pendulum is dependent on the length of the string holding the pendulum, L, and acceleration due to gravity, g. It is given mathematically as:

$T = 2\pi \sqrt{\frac{L}{g} }$

Let us make L the subject of the formula:

$T^2 = 4\pi ^2(\frac{L}{g}) \\\\\\\frac{L}{g} = \frac{T^2}{4\pi ^{2}} \\\\\\L = \frac{gT^2}{4\pi ^{2}}$

We are not told that the length of the string changes, hence, we can conclude that it is constant in both locations.

When the period of the pendulum is 2 s and the acceleration due to gravity is 9.8 $m/s^2$ , the length L is:

$L = \frac{9.8 * 2^2}{4 \pi^{2}}\\ \\\\L = 0.9929 m$

When the pendulum is moved to a new location, the period becomes 1.99782 s.

We have concluded that length is constant, hence, we can find the new acceleration due to gravity, $g_n$ :

$0.9929 = \frac{g_n * 1.99782^2}{4\pi^{2}} \\\\\\0.9929 = 0.1011 g_n$

Therefore:

$g_n = 0.9929/0.1011\\\\\\g_n = 9.821 m/s^2$

The difference between the new acceleration due to gravity, $g_n$ and the former acceleration due to gravity, g, will be:

$g_n - g$ = $9.821 - 9.8$ = $0.021 m/s^2$

The acceleration due to gravity differs by a value of $0.021 m/s^2$ at the new location.

7 0

3 years ago

Read 2 more answers

student drove to the university from her home and noted that the odometer on her car increased by 14.0 km. The trip took 18.0 mi

Darina [25.2K]

Answer:

a) Her average speed was 0.778 km/min

b) Her average velocity was 0.572 km/min

Explanation:

a) The average speed is given by the traveled distance over time:

average speed = distance / time

average speed = 14.0 km / 18.0 min = 0.778 km/min

b) The average velocity is given by the displacement over time. It is calculated as the variation of the position over time relative to the frame of reference. In this case, the origin of the frame of reference is located at the home of the student. The final position will be the university. Then:

velocity = (final position - initial position)/ time

velocity = 10.3 km / 18.0 min = 0.572 km/min

6 0

3 years ago