Question

Consider a block of mass 2 kg moving in the positive x-direction with a speed of 3 m/s at when it is at x=0. The block is subject to the net force shown in the plot below.

A) find the work done by the net force on the block as it moves from x=0 to x=2 m

B) find the speed of the block at x=2m

6 4 z (m) (In

Answer 1

from the graph the net work done is equal to area of the graph

W = area of recatngle side 2x2 + area of triangle of base 2 and height (6-2)

W = (2*2) + (0.5*2*4) = 8 J <<--------answer

++++++++++

from work energy theorem

work done = change in KE

W = 0.5*m*(v^2-u^2)

8 = 0.5*2*(v^2-3^2)

v = 4.123 m/s <<<---answer