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SOLUTIONS

1) F_NET = ma (Newton’s 2^nd Law of Motion)

a = F_NET/ m

= 10 N / 2kg

= 5 m/s² [in the direction of the applied horizontal force]

Velocity of the object after 2s is: 

V = a* D t

= (5)*(2)

= 10 m/s [in the direction of the applied force]

After 2s, the applied force is zero. According to Newton’s First Law of Motion, the object continues to move in the same direction with constant velocity of 10 m/s. Therefore, the object’s velocity after 4s is 10 m/s [in the direction of the applied force]

2) F = m₁a₁ (Newton’s 2^nd Law of Motion)

F = m₂a₂

F = (m₁ + m₂)a

F = ( F/2 + F/3 )* a

a = 6 F/ 5 F

a = 1.2 m/s² [in the direction of the applied force]

3) Directions: East (+)   West (-)    South (-)     North (+)

F_NET = ma (Newton’s 2^nd Law of Motion)

F_NETy:   F_y + F_g = F_y                                                                                        

F_NETx:   F_x – F_f = m* a                                                                         

(10)(Cos 37) – (5.2) = 2 * a

a = (2.8) / 2

= 1.4 m/s² [in the direction of the applied force]

 4) F_NET = m_system*a (Newton’s 2^nd Law of Motion)

10 N = (2kg + 3kg)* a 

a = 2 m/s² [in the direction of the applied force]

Let’s look at the forces acting on each object

F_NET = m₁ a

F_NET1y: 0

F_NET1x:         F - T = m₁a

                  10 – T = (2) (2)

T = 6 N

5) F_NET = m_system*a (Newton’s 2^nd Law of Motion)                                                                          Directions:

      20 = (4+1)* a                                                                                                                               left (-) � right(+)

       a = 4m/s² [forward]                                                                                                                (backward)     (forward)

Let T be the backward force of m₂ on m₁

Then,

F_NET1 = m₁ a

F_NET1y: 0

F_NET1x:    F – T = m₁* a

                20 – T = (4)(4)

T = 4 N

The backward force of m₂ on m₁ is 4 N

6)  F₁ = m*a₁

F₂ = m*a₂   (Newton’s 2^nd Law of Motion)

 D d = V₁ Dt + � aD t²

D d₁= X                      D t = t

X = 0 + � a₁t²

a₁ = 2X / t²

D d₂= 2X     D t = 2t

2X = 0 + � a₂(2t)²

a₂ = X / t²

a₁ / a₂ = 2

F₁ / F₂ = ma₁ / ma₂

\ F₁ / F₂ = 2

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