In a physics laboratory experiment, a 6.00 kg box is pushed across a flat table by a horizontal force F.
A. If the box is moving at a constant speed of 0.35 m/s and the coefficient of friction is 0.12. What is the magnitude of F?
B. What is the magnitude of F if the box is speeding up with constant acceleration of 0.18 m/s2?
A. To determine the magnitude of the force F when the box is moving at a constant speed of 0.35 m/s, we need to consider the forces acting on the box.
1. The force applied horizontally, F (pushing force).
2. The force of friction acting in the opposite direction to the motion, given by the equation:
frictional force = coefficient of friction x normal force
The normal force is equal to the weight of the box, which can be calculated using the equation:
weight = mass x gravitational acceleration
Since the box is not accelerating vertically (constant speed), the normal force is equal to the weight.
Now let's calculate the value of the force F.
1. Calculate the weight of the box:
weight = mass x gravitational acceleration
= 6.00 kg x 9.8 m/s^2
= 58.8 N
2. Calculate the force of friction:
frictional force = coefficient of friction x normal force
= 0.12 x 58.8 N
= 7.056 N
3. The force F is equal to the force of friction:
F = frictional force
= 7.056 N
Therefore, the magnitude of force F is 7.056 N.
B. To determine the magnitude of force F when the box is speeding up with a constant acceleration of 0.18 m/s^2, we need to consider the additional force required to overcome the frictional force and accelerate the box.
1. The force applied horizontally, F (pushing force).
2. The force of friction acting in the opposite direction to the motion, as calculated in part A.
Since the box is accelerating, the net force acting on the box is given by:
net force = frictional force + mass x acceleration
Now let's calculate the value of the force F.
1. Calculate the net force:
net force = frictional force + mass x acceleration
= 7.056 N + (6.00 kg x 0.18 m/s^2)
= 7.056 N + 1.08 N
= 8.136 N
2. The force F is equal to the net force:
F = net force
= 8.136 N
Therefore, the magnitude of force F is 8.136 N.
To find the magnitude of the force F in both scenarios, we need to consider the forces affecting the box: the force of friction and the force applied by the horizontal force F.
A. When the box is moving at a constant speed of 0.35 m/s, it means the net force acting on the box is zero (since the acceleration is zero). Therefore, the force F applied by the experimenter must counterbalance the force of friction to maintain this constant speed.
The force of friction can be calculated using the equation:
frictional force = coefficient of friction * normal force
The normal force is equal to the weight of the box, which can be calculated as:
weight = mass * gravitational acceleration
The gravitational acceleration is approximately 9.8 m/s^2.
Therefore, the normal force is:
normal force = 6.00 kg * 9.8 m/s^2
Once we have the normal force, we can calculate the frictional force:
frictional force = 0.12 * normal force
Since the box is moving at a constant speed, the force F must equal the frictional force. Thus:
F = frictional force
Substituting the values:
F = 0.12 * (6.00 kg * 9.8 m/s^2)
B. When the box is accelerating with constant acceleration of 0.18 m/s^2, it means the applied force F is greater than the frictional force. In this case, the net force acting on the box is equal to the mass of the box multiplied by the acceleration.
net force = mass * acceleration
Therefore, we have:
F - frictional force = mass * acceleration
Substituting the known values:
F - (0.12 * (6.00 kg * 9.8 m/s^2)) = 6.00 kg * 0.18 m/s^2
Solving this equation will give us the magnitude of the force F.