At the end of 3N force acts on the object during time interval of 1.5seconds with force acting towards right. A constant force of 4N to left is applied for 3seconds. What is the velocity at the end of the 3seconds ?

Answer:

v_f = -7.5 m / s

Explanation:

Let's analyze this exercise a little, two forces that act on a body for different time intervals are indicated, each force creates an impulse and since this is a vector quantity we must add in the form of vectors. The net momentum is

we assume that the direction to the right is positive

             I = I₁ + I₂

             I = F₁ Δt₁ - F₂ Δt₂

             I = 3  1.5 - 4  3

             I = -7.5   N s

now let's use the relationship between momentum and momentum, suppose the object starts from rest (vo = 0)

             I = Δp

             I = m (v_f - v₀)

             v_f = I / m

              v_f = -7.5 / m

to finish the calculation we must assume a mass m = 1 kg

              v_f = -7.5 m / s

the negative sign in the body is moving to the left

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