A particle of mass 0.40 kg moving initially with constant speed of 10m/s to the north is subject to a constant force of 8.0 N directed towards south for 30s. Take at that instant, the force is applied to be t = 0, and the position of the particle at that time to be x = 0, predict its position at t = -5s, 25s, 30s?
CLASS - XI PHYSICS (Laws of Motion)
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mass = 0.4kg
force = 8N
acceleration = 8/0.4 = 20 m/s²
Assume direction from south to north as positive,
at the instant when the force acts,
u = 10 m/s
a = -20 m/s²
x₀ = 0
x = x₀ + ut + 0.5at²
a)before t=0s, force was not acting
so 0 = x + ut
⇒ 0 = x + 10×(5)
⇒ x = -50m
So particle was 50m south of x=0 at t=-5s.
b)at t = 25s,
x = 10×25 - 0.5×20×25² = -6000 m
So particle is at 6000m south of x=0 at t=25s.
c)at t = 30s,
x = 10×30 - 0.5×20×30² = -8700 m
So particle is at 8700m south of x=0 at t=30s.
force = 8N
acceleration = 8/0.4 = 20 m/s²
Assume direction from south to north as positive,
at the instant when the force acts,
u = 10 m/s
a = -20 m/s²
x₀ = 0
x = x₀ + ut + 0.5at²
a)before t=0s, force was not acting
so 0 = x + ut
⇒ 0 = x + 10×(5)
⇒ x = -50m
So particle was 50m south of x=0 at t=-5s.
b)at t = 25s,
x = 10×25 - 0.5×20×25² = -6000 m
So particle is at 6000m south of x=0 at t=25s.
c)at t = 30s,
x = 10×30 - 0.5×20×30² = -8700 m
So particle is at 8700m south of x=0 at t=30s.
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