p07_040.pdf

40. The acceleration is constant, so we may use the equations in Table 2-1. We choose the direction of

motion as + x and note that the displacement is the same as the distance traveled, in this problem. We

designate the force (assumed singular) along the x direction actingon the m =2 . 0 kgobject as F .

(a) With v 0 = 0, Eq. 2-11 leads to a = v/t .AndEq.2-17gives∆ x = 2 vt Newton’s second law yields

the force F = ma . Eq. 7-8, then, gives the work:

W = F ∆ x = m v

2 mv 2

as we expect from the work-kinetic energy theorem. With v = 10 m/s, this yields W = 100 J.

(b) Instantaneous power is deﬁned in Eq. 7-48. With t =3 . 0 s, we ﬁnd

P = Fv = m v

P = Fv =33W .

2 vt = 1

v =67W .