Question

A body of mass 200g is moving horizontally with a velocity of 5m/s along the positive X- direction. At time t=0, when the body is at x=0, a constant force of 0.4N directed along the negative x- direction is applied to the body for 10s. What is the position(x) of the body at t=30s ?​

Answer 1

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Answer 2

Answer:

m= 200g = 0.2kg

initial velocity (u) = 5m/s

F = -0.4N (negative x axis direction

F =ma

a =F/m

0.4/0.2= 2m/s^2

since force is applied only for 10s

so a=-2m/s^2 for 10s

from

s = ut + 1/2 at^2

s = 5(10) + 1/2(-2)(100)

s= 50-100

s= -50m ----(1)

now velocity at t=10

v=u+at

v=5 - 2(10)

v= -15m/s

now u = -15m/s t=20s a=0(as force stopped acting f=0)

s=ut +1/2at^2

s= -15(20)

s= -300m----(2)

position of body at t=30s is (1) + (2)

= -50 + (-350)

= -350m