60. The acceleration is constant, so we may use the equations in Table 2-1. We choose the direction of motion as +xand note that the displacement is the same as the distance traveled, in this problem. We designate the force (assumed singular) along the xdirection acting on the m= 2.0 kg object as F.(a) With v0= 0, Eq. 2-11 leads to a= v/t. And Eq. 2-17 gives Δxvt12=. Newton’s second law yields the force F= ma. Eq. 7-8, then, gives the work:
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