How Does Inclination and Friction Affect Work and Energy on a Microwave Oven?

[Edwardo_Elric]

Homework Statement

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.

Homework Equations

W = F * x
U = mgy
K = 1/2(mv^2)
U1 + K1 + Wother = U2 + K2

The Attempt at a Solution

I think there are some typo errors from the back of the book:
a.) What is the work done on the oven by the force F?
W = F * x
W = 120N * (14.0cos(37)) <<<< (x component)
W = 1341.71 J

(the answer at back of book is 1690J)

b.) What is the work on the oven by the friction force?
Ff = ukn = 0.25(12.0kg)(9.8m/s^2) = 29.4N

Wf = Ff ( x)
Wf = (29.4N * 14.0cos(37))
Wf = 328.72J = 329J
(the answer at back of book is 329J)

c.) Compute the increase in potential energy for the oven.
U2 = mgy2 = (12.0kg)(9.8m/s^2)(14.0sin(37)) = 990.8J = 991J

(the answer at back of book is 991J)

d.) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy.

U1 + K1 + Wother = U2 + K2
0 + 0 + (WF - Wf ) = U2 + K2
1341.71 J - 329J - 991J = K2
K2 = 21.71J

(the answer at back of book is 360J)

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.
F - Ff = 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)
Vf = 14.5396m/s

K = 1/2(mv^2)
K = 1/2(12.0kg)(14.5396m/s)
K = 87.238J <<< not same as d i don't know why
(the answer at back of book is a = 0.699m/s^2: K = 360J)

 

[bel]

a) W= {\int_a}^b F ds, which is to say you don't only take the x component. (So the book did have a typing error here, but the error is less than one percent.)
b) The normal force here is not the same as the gravitational force.
c) E_K = E_{total} - (E_P + W)
d) Remember that forces are vectors and they point in different directions, so \sum F results in a vector pointing downhill (i.e., direction of acceleration).

 

[Edwardo_Elric]

a.) W = 14.0m(120N)
W = 1680N

b.) Ff = uk(mgcos(theta))
= 0.25(12.0kg(9.8m/s^2)(cos(37))
= 23.4799N
Wf = Ff * d
Wf = 23.4799N(14.0m)
Wf = 329N

c.) didnt follow your formula but increase in potential energy:
U2 = (12.0kg)(9.8m/s^2)(14sin(37))
U2 = 991J
same as B.O.B

d.) U1 + K1 + Wother = U2 + K2
0 + 0 + (WF - Wf ) = U2 + K2 (U1 and K1 are zero ? because the increase of K2 is required )
1680 J - 329J = 991J + K2
K2 = 360J
yup - same ans at Back

e.) F - Ff = ma
(120N - 23.4799N) / 12.0kg = a
a= 8.043m/s^2
(not the same as back of book)

vf^2 = v0^2 + 2ad
vf^2 = 2( 8.043m/s)(14.0m)
Vf = 15.004m/s

K2 = 1/2(mv^2)
K2 = 1/2(12.0kg( 15.004m/s)^2)
K2 = 1350.72J
not also the same

kindly check this

 

[learningphysics]

a) and b) should be in J... other than that a),b),c) and d) are good.

For part e) you need to consider the component of gravity acting along the ramp... for the acceleration along the ramp, you need the net force along the ramp...

 

[Edwardo_Elric]

ok
e.) Fnet = ma
F - Ff - wsin(theta) = ma
(120N - 23.4799N - ((9.8*12)(sin(37))/12kg = a
a = 25.75N / 12kg
a = 2.1455m/s^2

vf^2 = 2(2.1455m/s^2)(14.0m)
vf = 7.75m/s

K2 = 1/2(12.0kg)(7.75m/s)^2
K2 = 360.375J

thanks

 

[learningphysics]

ok
e.) Fnet = ma
F - Ff - wsin(theta) = ma
(120N - 23.4799N - ((9.8*12)(sin(37))/12kg = a
a = 25.75N / 12kg
a = 2.1455m/s^2

vf^2 = 2(2.1455m/s^2)(14.0m)
vf = 7.75m/s

K2 = 1/2(12.0kg)(7.75m/s)^2
K2 = 360.375J

thanks

no prob. It looks good... But you said the back of the book says a=0.699m/s^2 ?

 

[Edwardo_Elric]

yes...
the b.o.b. is 0.699m/s^2
is that a typo?

 

[learningphysics]

yes...
the b.o.b. is 0.699m/s^2
is that a typo?

Yeah, I think so.