High School

A 12.0 kg microwave oven is pushed 14.0 m up the sloping surface of a loading ramp inclined at an angle of 37 degrees above the horizontal by a constant force [tex]F[/tex] with a magnitude of 120 N, 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 [tex]F[/tex]?

b.) What is the work done 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 [tex]\sum{F} = ma[/tex] to calculate the acceleration of the oven. Assuming the oven is initially at rest, calculate the oven's speed after traveling 14.0 m. Compute the increase in the oven's kinetic energy from this, and compare it to the answer you got in part (d).

Answer :

a) The work done on the oven by the force F is 1341.7.

b) The work on the oven by the friction force is 329J.

c) The increase in potential energy for the oven is 991J.

d) The increase in oven's kinetic energy is 21.71J.

e) K.E is 87.23 j

a)

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

[tex]W = 120 N \times (14.0 cos(37))[/tex]

[tex]W = 120 N \times (14.0 cos(37))[/tex]

[tex]W = 1341.71[/tex]

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

[tex]F_f=0.25*12\times9.8[/tex]

[tex]= 29.4 N[/tex]

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

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

[tex]W_f= 328.72 J = 329 J[/tex]

c) increase in the internal energy

[tex]U_2 = mgh[/tex]

[tex]= 12\times9.81\times14sin(37)[/tex]

[tex]=[/tex] [tex]991 J[/tex]

d) the increase in oven's kinetic energy

[tex]U_1 + K_1 + W_other = U_2 + K_2[/tex]

[tex]1341.71 J - 329 J - 991 J = K_2[/tex]

[tex]K_2 = 21.71 J[/tex]

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

[tex](120N - 29.4N ) / 12.0kg = a[/tex]

[tex]a = 7.55[/tex]

[tex]vf^2 = v0^2 + 2ax[/tex]

[tex]vf^2 = 2(7.55m/s)(14.0m)[/tex]

[tex]V_f = 14.5396m/s[/tex]

[tex]K = 1/2(mv[/tex]

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

[tex]K = 87.238J[/tex]

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




Other Questions