Question

A 0.414 kg object is at rest at the origin of a coordinate system. A 3.17 N force in the positive x direction acts on the object for 2.98 s. What is the velocity at the end of this interval?

b. At the end of this interval, a constant force of 4.04 N is applied in the negative x direction for 4.17 s. What is the velocity at the end of the 4.17 s?

Answer 1

from, impulse = Force( F) * time(t) = m* ( v- u)

F*t = v-u              ----------------1                            ( v = final velocity, u = initial velocity)

since object is at rest, so u = 0 m/s

from eq, 1,            3.17*2.98 = 0.414 * ( V - 0)

V = 22.8 m/s

b) at the end of 2.98 sec, object have velocity = 22.8 m/s

so u = 22.8 m/s

F'*t' = m* ( V' - u)

4.04*4.17 = 0.414 * ( V' - ( - 22.8)                   ( -ve sign is used , because initial and final velocity are in opposite direction. final velocity is in -x direction & initial was in +x direction)

40.692 = V' + 22.8

V' (velocity at the end of 4.17 sec) = 17.89m/s