(a) The net force
"F_{net}=F-F_f=F-\\mu mg=275-0.358\\cdot 92\\cdot 9.8=-47.8N"
Opposite to the motion
(b) The net work done
"W_{net}=F_{net}\\cdot \\Delta x=-47.8\\cdot 0.65=-31.1 J"
(c) From the Work-Energy Theorem we have
"W=\\frac{1}{2}(mv^2_f-mv^2_i)"
"v_f=\\sqrt{ \\frac{2W+mv^2_i}{m}}=\\sqrt{ \\frac{2\\cdot (-31.1)+92 \\cdot 0.85^2}{92}}\\approx0.22 m\/s"
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