Summary of Equations

F = ma
a = F/m

Two objects (A and B) with same mass:
a_B/a_A = F_B/F_A

Two objects acted upon by the same force:
a_B/a_A = m_A/m_B

Acceleration of gravity:
g = 9.8 m/s²
Object falling from rest:
v = gt
D = (1/2)gt²

Weight = mg

Problem 1

F = 24 N, m = 3 kg
a = F/m, a = 24/3 = 8 m/s²

Problem 2

F = 2.5 N, m = 1000 kg
a = F/m = 2.5/1000 = 0.0025 m/s²
D = (1/2)at² = (1/2)(0.0025)(10)² = (1/2)(0.0025)(100) = 0.125 m.

Problem 3

m = 1000 kg, a = 0.01 m/s²
F = ma = (1000)(0.01) = 10 N.

Problem 4

a = 20 m/s², F = 6 N.
If F is changed to 18 N, it is multiplied by 3. Hence acceleration is also multiplied by 3:
a = 60 m/s².

F_A = 6 N, a_A = 20 m/s²
F_B = 9 N, a_B = ?
a_B/a_A = F_B/F_A
a_B/20 = 9/6 = 1.5
a_B = 20 X 1.5 = 30 m/s².

Problem 5

a = 6 m/s². Three girls exert three times as much force. Hence the acceleration is 3 times as much.
a = 18 m/s².

Problem 6

m_A = 1200 kg, a_A = 2 m/s²
m_B = 4800 kg, a_B = ?
a_B/a_A = m_A/m_B
a_B/2 = 1200/4800 = 0.25
a_B = 2 X 0.25 = 0.5 m/s².

Problem 7

t = 3 s
D = (1/2)gt² = (1/2)(9.8)(3)² = (1/2)(9.8)(9) = 44.1 m.

t = 6 s. Time is twice as much as previous example. Hence distance is multiplied by 2² = 4.
Distance = 4 X 44.1 = 176.4 m.

Problem 8

t = 0.2 s
v = gt = (9.8)(0.2) = 1.96 m/s

Problem 9

t = 15 s
v = gt = (9.8)(15) = 147 m/s
D = (1/2)gt² = (1/2)(9.8)(15)² = (1/2)(9.8)(225) = 1,102.5 m

Problem 10

m_A = 1.5 kg, a_A = 3.2 m/s²
m_B = 6 kg, a_B = ?
a_B/a_A = m_A/m_B
a_B/3.2 = 1.5/6 = 0.25
a_B = (3.2)(0.25) = 0.8 m/s²

Problem 11

m = 2 kg, F = 10 N
a = F/m = 10/2 = 5 m/s²
v = at = (5)(3) = 15 m/s
D = (1/2)at² = (1/2)(5)(3)² = (1/2)(5)(9) = 22.5 m

Problem 12

In case A the mass is just that of the wagon: m_A = 15 kg
In case B the mass is that of the wagon plus 6 kg: m_B = 15 + 6 = 21 kg.
a_A = 2.5 m/s², a_B = ?
a_B/a_A = m_A/m_B
a_B/2.5 = 15/21 = 0.714
a_B = (2.5)(0.714) = 1.79 m/s²

Problem 13

Weight = mg = (100)(9.8) = 980 N