Question

For a three-second period of time, a 50 kg wooden crate slides across the concrete floor at exactly 5 m/s. The coefficient of kinetic friction between the wooden crate and the concrete floor is 0.2. What is the net force of this crate during this three-second period?

A) 500 N
B) 50 N
C) 0 N
D) 100 N

Answer 1

The net force acting on the wooden crate during the three-second period is 50 N. Thus the correct opinion id B.

Step-by-step explanation:

To calculate the net force acting on the wooden crate, we can use Newton's second law, which states that
`\( \text{Force} = \text{Mass} * \text{Acceleration} \)`. The acceleration (a) can be determined using the equation of motion
`\( \text{Final Velocity} = \text{Initial Velocity} + (a * \text{Time}) \)`. The crate is initially at rest, so the initial velocity is 0 m/s. Rearranging the equation, we get
`\(a = \frac{\text{Final Velocity}}{\text{Time}}\)`, and substituting the values
`(\(a = \frac{5 \, \text{m/s}}{3 \, \text{s}} = 1.67 \, \text{m/s}^2\))`. Now, using Newton's second law,
`\( \text{Force} = \text{Mass} * \text{Acceleration} = 50 \, \text{kg} * 1.67 \, \text{m/s}^2 = 83.5 \, \text{N}\)`. However, this force is the force applied, and to find the net force, we need to consider the frictional force. The frictional force is
`\( \text{Frictional Force} = \text{Coefficient of Friction} * \text{Normal Force} \)`. The normal force is equal to the weight of the crate
`(\( \text{Weight} = \text{Mass} * \text{Gravity} = 50 \, \text{kg} * 9.8 \, \text{m/s}^2 = 490 \, \text{N}\)), so \( \text{Frictional Force} = 0.2 * 490 \, \text{N} = 98 \, \text{N}\)`. The net force is then
`\( \text{Force} - \text{Frictional Force} = 83.5 \, \text{N} - 98 \, \text{N} = -14.5 \, \text{N}\)`. However, the direction of the force matters, and since the force applied is in the opposite direction of the frictional force, we consider the net force as positive. Therefore, the net force is 14.5{N).

The calculation shows that the net force acting on the wooden crate during the three-second period is
`\(14.5 \, \text{N}\)`, and the correct option is not listed among the choices provided. If none of the given options matches the calculated value, it's advisable to double-check the calculation or consider the possibility of an error in the provided answer choices.