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3 years ago

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18. A person pulls horizontally with a force of 64 N on a 14-kg box. There is a force of friction between the box and the floor

of 36 N. Find the acceleration of the box in m/s2.

1 answer:

MA_775_DIABLO [31]3 years ago

4 0

since force=massxacceleration
therefore acceleration=force/mass
the resultant force=the acting force on the object-the frictional force
the resultant force=64-36=28N
therefore the acceleration=28/14=2m/sec^2

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4 years ago

Which is an example of transforming potential energy to kinetic energy? Select two options.

Rasek [7]

<u>Complete Question:</u>

Which is an example of transforming potential energy to kinetic energy? Select two options.

changing thermal energy to electrical energy

changing chemical energy to thermal energy

changing nuclear energy to radiant energy

changing radiant energy to electrical energy

changing mechanical energy to chemical energy

<u>Correct Answer:</u>

The examples of transforming potential energy to kinetic energy are changing chemical energy to thermal energy and changing nuclear energy to radiant energy.

<u>Explanation:</u>

As stated by the conservation of energy law, any form of energy is usually transferred to another form. The basic kinds of energy is potential and kinetic energy. Potential energy is the energy stored for the objects at rest and kinetic energy is the energy utilized by the objects for motion.

So in the given options, chemical energy is the energy stored by the chemical bonds to make a stable compound and that energy is converted to thermal energy when the bonds get broken. So the stored energy or the energy required to keep the bonds intact is chemical energy and it is thus a form of potential energy.

And when these bonds get broken, the electrons use the thermal energy released by this breakage as their kinetic energy. So one form of transforming potential energy to kinetic energy is by changing chemical energy to thermal energy.

Similarly, the nuclear energy is exhibited by the elementary particles in an atom. So it is similar to potential energy and the radiant energy is released whenever there is an excitation. So the radiant energy will be similar to kinetic energy.

Thus, the changing of chemical energy to thermal energy and the changing of nuclear energy to radiant energy are the examples of transforming potential energy to kinetic energy.

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3 years ago

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

4 years ago

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nata0808 [166]

Answer:

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Explanation:

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3 years ago