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 #66014, Physics / Mechanics | Relativity

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?

Solution:

Let $m$ be the truck's mass, $F_{fr}$ — average frictional force, $F_{t}$ — force generated by the truck, $v_{1}$ — initial truck's speed, $v_{2}$ — its final speed, $D$ — distance, $a$ — acceleration of the truck, and $t$ — time spent for increasing the speed.

According to Newton's second law of motion $\vec{F} = \vec{F}_t + \vec{F}_{fr} = m\vec{a}$.

In scalar form we may write it like $F_{t} - F_{fr} = ma$, because frictional force is contradirectional to the force generated by truck. So, $F_{t} = F_{fr} + ma$.

Let's determine truck's acceleration.

For uniformly accelerated motion

$$v _ {2} = v _ {1} + a t$$ $$D = v _ {1} t + \frac {a t ^ {2}}{2}$$

From $1^{\mathrm{st}}$ equation $t = \frac{v_2 - v_1}{a}$ and then $= v_{1}\frac{v_{2} - v_{1}}{a} +\frac{a\left(\frac{v_{2} - v_{1}}{a}\right)^{2}}{2} = v_{1}\frac{v_{2} - v_{1}}{a} +\frac{(v_{2} - v_{1})^{2}}{2a} = \frac{v_{2}^{2} - v_{1}^{2}}{2a}.$

Finally, $a = \frac{v_2^2 - v_1^2}{2D}$ and $F_{t} = F_{fr} + m\frac{v_{2}^{2} - v_{1}^{2}}{2D}$.

$$F _ {t} = 8 0 0 + 2 0 0 0 \frac {3 5 ^ {2} - 2 5 ^ {2}}{2 \cdot 5 0 0} = 2 0 0 0 N$$

Answer:

2000 N

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