Question

A railroad car moves under a grain elevator at a constant speed of $\mathbf{3}\mathbf{.}\mathbf{20}\frac{\mathbf{m}}{\mathbf{s}}$. Grain drops into the car at the rate of$\mathbf{540}\frac{\mathbf{kg}}{\mathbf{min}}$. What is the magnitude of the force needed to keep the car moving at constant speed if friction is negligible?

Short answer

Magnitude of force needed to keep the car moving at constant speed is 28.8 N

Step-by-step solution

Step 1: Given

Car speed is,$v_{rel}=3.20\frac{\mathrm{m}}{\mathrm{s}}$

Grain’s Massrate is,
$$\begin{gathered} R=540\frac{\mathrm{kg}}{\min } \\ =\frac{540}{60}\frac{\mathrm{kg}}{\sec } \end{gathered}$$

Determine the concept

Use the formula for force in terms of mass rate and relative velocity, find the magnitude of force needed to keep the car moving at constant speed.

Formula is as follow:

$F=R{v}_{rel}$

Here, F is force, R is mass rate and $v_{rel}$is relative velocity.

Determining the magnitude of force needed to keep the car moving at constant speed

Magnitude of force needed to keep the car moving is,

$$\begin{gathered} F=R\times v_{rel} \\ F=\frac{540}{60}\times 3.2 \\ F=28.8\mathrm{N} \end{gathered}$$

Hence, themagnitude of force needed to keep the car moving at constant speed is

28.8 N

Therefore, the additional force required to keep moving the object can be found using the mass rate of the other object and relative velocity.