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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)
    100 N
  • c)
    50 N
  • d)
    0 N
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
For a three second period of time, a 50 kg wooden crate slides across ...
  • Drawing a free-body diagram of the crate in motion will elucidate the forces on the crate.
  • The forces present in the vertical direction are the normal force and the gravitational force; these forces cancel out. The forces present in the horizontal direction are the force due to kinetic friction and another applied force in the opposite direction of the friction since there is a constant velocity that does not dissipate due to the friction.
  • Since there is a constant velocity, there is no net acceleration and no net force.
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Most Upvoted Answer
For a three second period of time, a 50 kg wooden crate slides across ...
Net force is the vector sum of all the forces acting on an object. In this case, the net force can be calculated by considering the force of friction and the force applied to the crate.

The force of friction can be determined using the equation:

Frictional force = coefficient of friction * normal force

The normal force is equal to the weight of the crate, which can be calculated using the equation:

Weight = mass * acceleration due to gravity

The force applied to the crate can be calculated using the equation:

Force = mass * acceleration

Given that the mass of the crate is 50 kg, the coefficient of kinetic friction is 0.2, and the velocity is 5 m/s, we can calculate the net force as follows:

1. Calculate the weight of the crate:
Weight = mass * acceleration due to gravity
Weight = 50 kg * 9.8 m/s^2
Weight = 490 N

2. Calculate the force of friction:
Frictional force = coefficient of friction * normal force
Frictional force = 0.2 * 490 N
Frictional force = 98 N

3. Calculate the force applied to the crate:
Force = mass * acceleration
Force = 50 kg * 5 m/s^2
Force = 250 N

4. Calculate the net force:
Net force = force applied - force of friction
Net force = 250 N - 98 N
Net force = 152 N

Therefore, the net force acting on the crate during this three-second period is 152 N. Since none of the given options match this value, the correct answer would be option 'D', which states that the net force is 0 N. However, it should be noted that this answer is incorrect based on the calculations.
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