A 2.2kg toy train is confined to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origin moving at a speed of 1.6m/s in the +x direction, and continues until it reaches a position 7.5m down the track from where it started. During its Journey, it experiences a force pointing in the same direction as the vector 0.61 + 0.81, with magnitude initially 2.8N and decreasing linearly with its x-position to 0 N when the train has finished its journey. a. Calculate the work done by this force over the entire journey of the train. b. Find the speed of the train at the end of its journey.

W = 10.51 J , v = 3.34 m / s

Explanation:

To find the work we use its definition

      W = F .dx

the bold are vectors.  In this case, as the force is in the direction of movement, the product will become algebraic.

       W = F dx

the force that is in the direction of the vector from the origin to the point (0.61, 0.81) this force can be decomposed into its x and y components, where only the x component does work, this component has an initial value of 2, 8 N in position xo = 1.6 and decreases linearly to zero in elections x = 7.5 m

with these data we can find the linear equation of this force

         F = m x + b

the slope is

         m = (2.8 -0) / (0 - 7.5)

         m = -0.373 N / m

we find the constant with the starting point

         2.8 = -0.373 0 + b

         b = 2.8 m

the equation of force remains

         F = -0.373 x + 2.8

we substitute in the work equation and integrated

      W = ∫(-0.373 x + 2.8) dx

       W = -0.373 x² / 2 + 2.8 x

We evaluate between the lower limits x = 0 m and upper x = 7.5 m

        W = -0.373 (7.5²/2 - 0) + 2.8 (7.5 -0)

        W = -10.49+ 21

        W = 10.51 J

To find the speed of the train, let's use the relationship between work and the variation of kinetic energy.

       W = ΔK

       W = ½ m v² - ½ m v₀²

       v = √ (v₀ + 2W / m)

       v = √ (1.6 + 2 10.51 / 2.2)

       v = 3.34 m / s