# Vertical motion

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#1
A stone of mass m is released from rest on the surface of a tank of water of depth d. During the motion, the water exerts a constant resisting force of magnitude R. The stone takes t seconds to reach the bottom of the tank. Show that R = m(g - 2d/t^2). mechanics 1 douglas quadling
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1 month ago
#2
(Original post by agbotse.joana)
A stone of mass m is released from rest on the surface of a tank of water of depth d. During the motion, the water exerts a constant resisting force of magnitude R. The stone takes t seconds to reach the bottom of the tank. Show that R = m(g - 2d/t^2). mechanics 1 douglas quadling
Have you read the posting guidelines?

Don't just "throw" a question without telling us what is your difficulty and expect others to give you a solution.
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#3
(Original post by Eimmanuel)
Have you read the posting guidelines?

Don't just "throw" a question without telling us what is your difficulty and expect others to give you a solution.
I don't understand the question. I have spent several hours trying to solve but I am unable to get the answer.
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1 month ago
#4
(Original post by agbotse.joana)
I don't understand the question. I have spent several hours trying to solve but I am unable to get the answer.
What have you tried?
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#5
well I guess I have solved it but I am not too sure. But here is the solution:
F = ma
F = mg
g = F/m
F - R = ma
F = mg
mg - R = ma
mg - ma = R
R = m(g - a)
To find a:
s = ut + 1/2at^2
u = 0 (from rest where the stone was released)
d = 0 + 1/2at^2
d = 1/2at^2
2d = at^2
a = 2d/t^2
hence mg - ma = R
m(g - a) = R
Therefore: R = m(g - 2d/t^2).
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1 month ago
#6
(Original post by agbotse.joana)
well I guess I have solved it but I am not too sure. But here is the solution:
F = ma
F = mg
g = F/m
F - R = ma
F = mg
mg - R = ma
mg - ma = R
R = m(g - a)
To find a:
s = ut + 1/2at^2
u = 0 (from rest where the stone was released)
d = 0 + 1/2at^2
d = 1/2at^2
2d = at^2
a = 2d/t^2
hence mg - ma = R
m(g - a) = R
Therefore: R = m(g - 2d/t^2).
It seems that you "like" to reuse the same symbol to represent different physical quantities. This is unacceptable.

Suggested steps:
Fnet = ma
W - R = ma
mg - R = ma
a = (mg - R)/m

I am good with the following steps.

(Original post by agbotse.joana)
To find a:
s = ut + 1/2at^2
u = 0 (from rest where the stone was released)
d = 0 + 1/2at^2
d = 1/2at^2
2d = at^2
a = 2d/t^2
hence mg - ma = R
m(g - a) = R
Therefore: R = m(g - 2d/t^2).
0
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