2211()(2)fiKmvvmaxmaxΔ=− =Δ= Δwhere we have used 2vvax=+Δfrom Table 2-1. From Fig. 7-27, we see that (0 30) J30 JKΔ=−=−when5 mxΔ=+. The acceleration can then be obtained as 2(30J)0.75 m/s .(8.0 kg)(5.0 m)KamxΔ−===−ΔThe negative sign indicates that the mass is decelerating. From the figure, we also see that when 5 mx=the kinetic energy becomes zero, implying that the mass comes to rest
This is the end of the preview. Sign up
access the rest of the document.