Question

mass of 2kg moves horizontally along x axis under the action of a force in terms of time, given as following: F(t) = b sin wt , where t time in seconds , b and w are constants

1) find the impulse during t1=0 to t2=2
if the mass starts motion from rest at x=0
(Show in integration IN DETAIL)

2) find its velocity as function of time
(IN DETAIL)

3) find its position as function of time
(IN DETAIL)

Answer 1

1) according to newton's second law

F = dp/dt

dp = F dt

taking integral both sides we have

p = b sin wt * dt from t = 0 to t = 2

p = -b/w cos wt from t= 0 to t = 2

p = -b/w (cos 2t - cos 0)

= b/w(1 - cos 2t)

2) p = mv

mv = b/w ( 1 - cos 2t)

2v = b/w(1- cos 2t)

v = b/2w ( 1 - cos 2t)

3) v = dx/dt

dx/dt = b/2w ( 1- cos 2t)

dx = b/2w ( 1 -cos 2t) dt

integrating both sides we have

x = b/2w ( 1 -cos 2t) dt

= b/2w ( t - sin 2t / 2) + C

at t= 0 , x = 0

0 = 0 - 0 + C

C = 0

x = b/4w (2t - sin 2t)