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A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horizontal, by a constant force F with a magnitude of 120N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.25.
a.) What is the work done on the oven by the force F?
b.) What is the work on the oven by the friction force?
c.) Compute the increase in potential energy for the oven.
d.) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy.
e.)Use summation{F} = ma to calculate the acceleration of the oven. Assuming the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 14.0m. Compute from this the increase in the oven's kinetic energ, and compare this answer to the one you got in part d.

Respuesta :

Manetho

Answer:

Answered

Explanation:

a) What is the work done on the oven by the force F?

W = F * x

W = 120 N * (14.0 cos(37))

<<<< (x component)

W = 1341.71

b) [tex]F_f=\mu_k N[/tex]

[tex]F_f=0.25\times12\times9.8[/tex]

= 29.4 N

[tex]W_f= F_f\times x[/tex]

[tex]W_f= 29.0\times 14 cos(37)[/tex]

W_f= 328.72 J = 329 J

c) increase in the internal energy

U_2 = mgh

= 12*9.81*14sin(37)

= 991 J

d) the increase in oven's kinetic energy

U_1 + K_1 + W_other = U_2 + K_2

0 + 0 + (W_F - W_f ) = U_2 + K_2

1341.71 J - 329 J - 991 J = K_2

K_2 = 21.71 J

e) F - F_f = ma

(120N - 29.4N ) / 12.0kg = a

a = 7.55m/s^2

vf^2 = v0^2 + 2ax

vf^2 = 2(7.55m/s)(14.0m)  

V_f = 14.5396m/s

K = 1/2(mv^2)

K = 1/2(12.0kg)(14.5396m/s)

K = 87.238J

shubhamchouhanvrVT

(a) Work done on the oven = 1341.71J

(b) Work done by the friction force =328.72J

(c) Increase in PE =991J

(d) Increase in KE = 21.71J

(e) Acceleration of the oven =7.55 [tex]\frac{m}{s^2}[/tex]  And KE =87.238J

what will be the value of KE, PE,  and acceleration for the oven?

a) The work done on the oven by the force is given by

[tex]W= F\times D[/tex]

[tex]W= 120N\times (14cos(37))[/tex]

It is a Horizontal component

[tex]W=1341.71J[/tex]

b) Now to calculate the work done by the friction force

[tex]F_f=\mu_kN\\\\F_f=0.25\times12\times9.8[/tex]

[tex]F_f=29.4N[/tex]

[tex]W_f=F_f\times D[/tex]

[tex]W_f=29\times14cos(37)[/tex]

[tex]W_f=328.72J[/tex]

c) increase in the Potential  energy

[tex]PE =mgh[/tex]

[tex]12\times 9.81\times14sin(37)[/tex]

[tex]PE=991J[/tex]

d) The increase in the oven's kinetic energy

[tex]PE_1+KE_!+W=PE_2+KE_2[/tex]

[tex]0+0(W_F-W_f)=PE_2+KE_2[/tex]

[tex]1341.71-329-991=KE_2[/tex]

[tex]KE_2=21.71J[/tex]

e) The acceleration of the oven

[tex]F-F_f=ma[/tex]

[tex]a=\dfrac{120-29.4}{12}= 7.55\frac{m}{s^2}[/tex]

[tex]a=7.55\dfrac{m}{s^2}[/tex]

Now to calculate KE we should know the velocity

[tex]V_f^2=V_0^2+2aD[/tex]

[tex]V_f^2=2(7.55)(14)[/tex]

[tex]V_f=14.53\frac{m}{s}[/tex]

Now KE will be given as

[tex]KE = \dfrac{1}{2} mv^2[/tex]

[tex]KE =\dfrac{1}{2} (12)(14.58)[/tex]

[tex]KE=87.238J[/tex]

Thus

(a) Work done on the oven = 1341.71J

(b) Work done by the friction force =328.72J

(c) Increase in PE =991J

(d) Increase in KE = 21.71J

(e) Acceleration of the oven =7.55 [tex]\frac{m}{s^2}[/tex]  And KE =87.238J

To know more about KE and PE follow

https://brainly.com/question/25959744

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