Subpart 3: Set up Newton's Second Law in the x-direction when the ice block is accelerating to the right with an acceleration a. (i) Write Newton's Second Law along the x-axis by adding all forces in the x-direction taking into account their signs (forces pointing to the right are positive and forces pointing to the left are negative) when the block is accelerating to the right, in terms of F, 0, and the force of static friction f, or the force of kinetic friction f whichever is applicable. DON'T forget the subscript on for f ΣF₂= ✔=ma (4) (ii) What expression of kinetic friction? O μ, N Hk N

Physics for Scientists and Engineers: Foundations and Connections
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ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter5: Newton's Laws Of Motion
Section: Chapter Questions
Problem 9PQ
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A contestant in a winter sporting event pushes a block of ice of mass m across a frozen lake as
shown in the figure. The coefficient of static friction between the block and ice is , and the
coefficient of kinetic friction is μ- 0 is the angle the force makes with the x-axis.
In this part, we are going to set-up Newton's second Law equations for the cases
(1) when the ice block just starts moving, and
(2) when it is accelerating to the right with an acceleration a.
All answers are symbolic. ALL ANSWERS ARE CASE-SENSITIVE.
Subpart 1: Newton's Second Law along the y-axis
(i) Write Newton's Second Law along the y-axis by adding all forces in the y-direction taking into
account their signs (forces pointing upwards are positive and downward are negative) in terms of the
normal force N, weight mg, F and 0.
In both scenarios, there is no acceleration along the y-direction, therefore, a, 0.
-Fsine+N-mg=may=0 (1)
ZF,=
(ii) Using (1) to solve for N
N = mg + Fsine (2)
Think: In (2) is N greater than the weight, less than the weight or equal to the weight?
Ⓒμ₂ N
O Mk N
Ө
Subpart 2: Set up Newton's Second Law in the x-direction when the ice block is stationary.
(i) Write Newton's Second Law along the x-axis by adding all forces in the x-direction taking into
accountr their signs (forces pointing to the right are positive and pointing to the left are negative)
when the block is not moving in terms of F, 0, and the force of static friction f.
DON'T forget the subscript on f.
If the block is not moving, then ax=0.
ZF, Fcos8-1,✔✔= max = 0 (3)
Y
(ii) What value of static friction should you use just before the block starts moving?
fs.max =
V
X
DON'T forget the subscript on f, or fk
ΣF₂=
✔-ma (4)
Ο με Ν
N
O μk N
Subpart 3: Set up Newton's Second Law in the x-direction when the ice block is accelerating to
the right with an acceleration a.
(ii) What expression of kinetic friction?
fk =
(i) Write Newton's Second Law along the x-axis by adding all forces in the x-direction taking into
account their signs (forces pointing to the right are positive and forces pointing to the left are
negative) when the block is accelerating to the right, in terms of F, 0, and the force of static friction
f, or the force of kinetic friction f whichever is applicable.
Transcribed Image Text:A contestant in a winter sporting event pushes a block of ice of mass m across a frozen lake as shown in the figure. The coefficient of static friction between the block and ice is , and the coefficient of kinetic friction is μ- 0 is the angle the force makes with the x-axis. In this part, we are going to set-up Newton's second Law equations for the cases (1) when the ice block just starts moving, and (2) when it is accelerating to the right with an acceleration a. All answers are symbolic. ALL ANSWERS ARE CASE-SENSITIVE. Subpart 1: Newton's Second Law along the y-axis (i) Write Newton's Second Law along the y-axis by adding all forces in the y-direction taking into account their signs (forces pointing upwards are positive and downward are negative) in terms of the normal force N, weight mg, F and 0. In both scenarios, there is no acceleration along the y-direction, therefore, a, 0. -Fsine+N-mg=may=0 (1) ZF,= (ii) Using (1) to solve for N N = mg + Fsine (2) Think: In (2) is N greater than the weight, less than the weight or equal to the weight? Ⓒμ₂ N O Mk N Ө Subpart 2: Set up Newton's Second Law in the x-direction when the ice block is stationary. (i) Write Newton's Second Law along the x-axis by adding all forces in the x-direction taking into accountr their signs (forces pointing to the right are positive and pointing to the left are negative) when the block is not moving in terms of F, 0, and the force of static friction f. DON'T forget the subscript on f. If the block is not moving, then ax=0. ZF, Fcos8-1,✔✔= max = 0 (3) Y (ii) What value of static friction should you use just before the block starts moving? fs.max = V X DON'T forget the subscript on f, or fk ΣF₂= ✔-ma (4) Ο με Ν N O μk N Subpart 3: Set up Newton's Second Law in the x-direction when the ice block is accelerating to the right with an acceleration a. (ii) What expression of kinetic friction? fk = (i) Write Newton's Second Law along the x-axis by adding all forces in the x-direction taking into account their signs (forces pointing to the right are positive and forces pointing to the left are negative) when the block is accelerating to the right, in terms of F, 0, and the force of static friction f, or the force of kinetic friction f whichever is applicable.
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