Question

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.

Answer 1

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) $F_f=\mu_k N$

$F_f=0.25\times12\times9.8$

= 29.4 N

$W_f= F_f\times x$

$W_f= 29.0\times 14 cos(37)$

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