Question

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.

Answer 1

Answer on Question #65541, Physics / Mechanics | Relativity |

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

Solution

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

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

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

Finally

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

Answer: 2000 N

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