Here
#m=2# object mass
#F = 30# applied force
#F_r = 5# friction force
#g = 9.8# gravity acceleration
#theta = pi/4# inclination angle
#d = 3# distance covered along the plane.
#a = ?# object acceleration after force application.
#v = ?# object's velocity after force application.
#s = ?# object's position along the plane after force application.
#t = ?# time.
The total work done on the object is responsible for the energy variation suffered by the object. Assuming that the initial point is a rest point, after the force application the object energy will be
#E_2 = d cos theta m g + 1/2m v_2^2#
The actuating forces projected over the inclined plane are
#F cos theta - m g sin theta-F_r = m a#
#s = s_0+v_0t+1/2 a t^2#so after #Delta t# we have
#d = 1/2 a Delta t^2# and #Delta t = sqrt((2d)/a)# and then
#v_2 = a Delta t = sqrt(2da)# so
#E_2 = d cos theta m g + m d a#
including the work done by the friction forces we have
#E_("total") = E_2 + F_r * d = d cos theta m g + m d a+F_r*d#
and substituting values
#E_("total") = 63.64# {J]
