Question #65493

A truck of mass 2000 kg moving on a highway experiences an average frictional force of 800 N. If its speed increases from 25 ms-1 to 35 ms-1 over a distance of 500 m, what is the force generated by the truck.
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Expert's answer

2017-02-23T11:59:05-0500

Answer on Question #65493, Physics / Mechanics | Relativity |

A truck of mass 2000kg2000\,\mathrm{kg} moving on a highway experiences an average frictional force of 800N800\,\mathrm{N}. If its speed increases from 25m/sec25\,\mathrm{m/sec} to 35m/sec35\,\mathrm{m/sec} over a distance of 500m500\,\mathrm{m}, what is the force generated by the truck.

Solution

m=2000kgFf=800Nv1=25m/secv2=35m/secL=500mF?\begin{array}{l} m = 2\,000\,\mathrm{kg} \\ F_f = 800\,\mathrm{N} \\ v_1 = 25\,\mathrm{m/sec} \\ v_2 = 35\,\mathrm{m/sec} \\ L = 500\,\mathrm{m} \\ F - ? \end{array}


We have a formula ma=FFfm \cdot a = F - F_f from the Newton's second law. Therefore F=Ff+maF = F_f + m \cdot a. We find the acceleration (a)(a) from this well-known equation:


a=v22v122L.a = \frac{v_2^2 - v_1^2}{2L}.


Finally


F=Ff+mv22v122L.F = F_f + m \frac{v_2^2 - v_1^2}{2L}.F=800+2000(352252)/1000=800+2(352252)=800+2600=2000(N).F = 800 + 2\,000 \cdot (35^2 - 25^2) / 1\,000 = 800 + 2 \cdot (35^2 - 25^2) = 800 + 2 \cdot 600 = 2\,000\,\mathrm{(N)}.


Answer: 2000 N

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