Question

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

Answer 1

Answer on Question 66882, Physics, Mechanics, Relativity

Question:

A truck of mass $2000\, kg$ moving on a highway experiences an average frictional force of $800\, N$. If it's 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's first apply the Newton's Second Law of Motion:

$$\sum F_x = m a_x,$$ $$F_{truck} - F_{fr} = m a,$$

here, $F_{truck}$ is the force generated by the truck, $F_{fr}$ is the force of friction, $m$ is the mass of the truck and $a$ is the acceleration of the truck.

Then, from this formula we can find the force generated by the truck:

$$F_{truck} = F_{fr} + m a \quad (1).$$

We can find the acceleration of the truck from the kinematic equation:

$$v_f^2 = v_i^2 + 2 a d,$$

here, $v_i$ is the initial speed of the truck, $v_f$ is the final speed of the truck, $a$ is the acceleration of the truck and $d$ is the distance. Then, we get:

$$a = \frac{v_f^2 - v_i^2}{2 d} = \frac{\left(35 \frac{m}{s}\right)^2 - \left(25 \frac{m}{s}\right)^2}{2 \cdot 500\, m} = 0.6 \frac{m}{s^2}.$$

Substituting the acceleration of the truck into the equation (1), we get:

$$F_{truck} = F_{fr} + m a = 800\, N + 2000\, kg \cdot 0.6 \frac{m}{s^2} = 2000\, N.$$

Answer:

$$F_{truck} = 2000\, N.$$

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