Buoyant Force & Gravity Force find Fnet (Newtons Second Law)

[tmschmidt24]

Homework Statement

A 4 kg book sinks a distance of 5m in the ocean starting from rest. The volume of the book is 0.00266666666666667 m^3. We assume the velocity of the Earth is zero during the whole process, and gravity and the bouyance force are the only interactions of the book. (Note: Take the density of sea water to be 1000 kg/m^3)
Solving the problem using Newton’s Seconds Law approach:
(Assume standard coordinates, centered on the initial position of the book.)
a) What is the net force acting on the book?
F_x^(net) = 0 N
F_y^(net) = ? N

b) What is the acceleration of the book?
a_x = 0 m/s^2
a_y = ? m/s^2

c) What is the final velocity of the book?
v_x^(f) = 0 m/s
v_y^(f) = ? m/s

d) What is the final speed of the book? |(v^->^f)| = ? m/s

Homework Equations

Fnet= F(gravity)+F(buoyancy)
Fnet=ma
F(gravity)=m*g
F(buoyancy)=p(fluid)V(sub)g
g=9.8 m/s^2

The Attempt at a Solution

F(buoyancy)=1000*0.00266666666666667*9.8= 26.13 N (since it is a force against it'd be negative)
F(gravity)=4*9.8=39.2 N
Fnet^y= 39.2-26.13= 13.07 N

It says my answer is wrong for Fnet^y. If I could just get the Fnet^y I can do the rest is there something I am missing?

 

[gneill]

Homework Statement

A 4 kg book sinks a distance of 5m in the ocean starting from rest. The volume of the book is 0.00266666666666667 m^3. We assume the velocity of the Earth is zero during the whole process, and gravity and the bouyance force are the only interactions of the book. (Note: Take the density of sea water to be 1000 kg/m^3)
Solving the problem using Newton’s Seconds Law approach:
(Assume standard coordinates, centered on the initial position of the book.)
a) What is the net force acting on the book?
F_x^(net) = 0 N
F_y^(net) = ? N

b) What is the acceleration of the book?
a_x = 0 m/s^2
a_y = ? m/s^2

c) What is the final velocity of the book?
v_x^(f) = 0 m/s
v_y^(f) = ? m/s

d) What is the final speed of the book? |(v^->^f)| = ? m/s

Homework Equations

Fnet= F(gravity)+F(buoyancy)
Fnet=ma
F(gravity)=m*g
F(buoyancy)=p(fluid)V(sub)g
g=9.8 m/s^2

The Attempt at a Solution

F(buoyancy)=1000*0.00266666666666667*9.8= 26.13 N (since it is a force against it'd be negative)
F(gravity)=4*9.8=39.2 N
Fnet^y= 39.2-26.13= 13.07 N

It says my answer is wrong for Fnet^y. If I could just get the Fnet^y I can do the rest is there something I am missing?

Hi tmschmidt24, Welcome to Physics Forums.

0.00266666666666667 m³ is a very strange value to be given for a volume. Are we to believe that the volume was measured to that many significant figures? The other given values are rather more limited, only one significant figure each. Could it be that the problem is testing your recognition/use of significant figures?

By the way, 0.00266666666666667 ≈ 1/375 . Just something I noticed, probably not important.

 

[tmschmidt24]

Thank you. The problem is copied word for word from web assign. It accepts 1 or 2 decimal points as an answer. I did try .0027 as well and came up with 26.46 for F(b) which make the F(net)y= 12.74, that did not work. It does not seem to car about Sig Figs do not matter on webassign unless stated. I thought maybe I was missing apart of the problem. I have sent an e-mail to my professor since it is due tonight at 11:59 CDST time tonight I hope that he will give me another extension.

 

[gneill]

Another possible "gotcha" is the assumption of coordinate system axes directions. Is your net y-direction force upwards or downwards? Which direction is positive?