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Chapter 5, Problem 2

Two horizontal forces act on a 4.0 kg chopping block that can slide over a frictionless kitchen counter, which lies in an xy plan
e
Answer (Part A)
0i + 0j m/s^2
Answer (Part B)
0i + 4.8j m/s^2
Answer (Part C)
3.25i + 0j m/s^2
How To Solve
~~~~~~~~~~~~~~
Refer to pages: (5.6, 3.6, 3.5)
Newton's Second Law
**********************
F net = Mass * Acceleration
Note: F net (sub x) = Mass * Accleration (sub x)
F net (Sub y) = Mass * Accleration (Sub y)
F net (sub x)
1) Realize that A (sub x) = 
Mass
2) Add the x components and divide by mass to find Accleration
Part A
X (6.5 + 6.5) / 4.0
= 0 m/s^2
Y (9.6 + 9.6) / 4.0
= 0 m/s^2
Part B
X (6.5 + 6.5) / 4.0
= 0 m/s^2
Y (9.6 +
9.6) / 4.0
= 4.8 m/s^2
Part C
X (6.5 +
6.5) / 4.0
= 3.25 m/s^2
Y (9.6 + 9.6) / 4.0
= 0 m/s^2

Chapter 5, Problem 6

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While two forces act on it, a particle is to move at the constant velocity
= (3.05 m/s)  (4.40 m/s). One of the forces is 1 = (2.5
3
Answer
2.53i + 6.42j N
How To Solve
~~~~~~~~~~~~~~
Refer to pages:
(3.5, 3.6, 5.6)
F Net is The sum of the forces
********************************
F (1) + F (2) = F net
Newtons 2nd Law
*****************
F net = Mass * Acceleration
1) Relize that Acceleration is 0 because velocity is constant.
Think about it, if you are moving at a constant speed and direction,
your moving, not accelerating.
Therfore:
F net = Mass * Acceleration
F net = Mass * 0
F net = 0
2)
F net = F (1) + F (2)
If one force is (2.53 N)i + (6.42 N)j Then
F (1)(subx) = 2.53 + "unknown" = 0
F (1)(suby) = 6.42 + "unknown" = 0
3) Solve algebraicly
Answer
2.53i + 6.42j Newtons

Chapter 5, Problem 14

A block with a weight of 8.6 N is at rest on a horizontal surface. A 3.3 N upward force is applied to the block by means of an a
t
Answer
5.3 N
How to solve
~~~~~~~~~~~~~~
Refer to pages: (5.6, 5.7, 5.8)
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 Winter '08
 Naik
 Physics, Acceleration, Force, Friction, Mass, natural force

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