A 10 kg wagon is pushed with a force of 7 newton for 1.5 seconds then with a force of 5 newton for 1.7 seconds and then with a force of 10 newton for 3 seconds in the same direction what is the change in velocity brought about???
Answers
m = mass of wagon = 10 kg
F = force applied = 7 N
t = time duration = 1.5 s
v₁ = initial velocity
v₂ = final velocity
using impulse -change in momentum equation
m (v₂ - v₁ ) = F t
10 (v₂ - v₁ ) = 7 x 1.5
v₂ - v₁ = 1.05 m/s
v₂ = v₁ + 1.05 eq-1
m = mass of wagon = 10 kg
F = force applied = 5 N
t = time duration = 1.7 s
v₂ = initial velocity
v₃ = final velocity
using impulse -change in momentum equation
m (v₃ - v₂ ) = F t
10 (v₃ - v₂ ) = 5 x 1.7
v₃ - v₂ = 0.85 m/s
v₃ - v₁ - 1.05 = 0.85
v₃ - v₁ = 1.9
v₃ = v₁ + 1.9 eq-2
m = mass of wagon = 10 kg
F = force applied = 10 N
t = time duration = 3 s
v₃ = initial velocity
v₄ = final velocity
using impulse -change in momentum equation
m (v₄ - v₃ ) = F t
10 (v₄ - (v₁ + 1.9) ) = 10 x 3
v₄ - v₁ - 1.9 = 3 m/s
v₄ - v₁ = 4.9
so change in velocity is 4.9 m/s