A man pulls a crate of mass 61.0 kg across a level floor. He pulls with a force of 178.0 N at an angle of 20.0� above the horizontal. When the crate is moving, the frictional force between the floor and the crate has a magnitude of 124.0 N.
If the crate starts from rest, how fast will it be moving after the man has pulled it a distance of 2.90 m?
Calculate the acceleration a using
F = M a a = F/M
F = 178.0 cos 20 - 124.0 N
The velocity after moving a distance X is sqrt(2aX)
To find the final speed of the crate, we need to use the laws of motion and the concept of work done.
1. First, let's calculate the net force acting on the crate. The net force is the vector sum of the applied force and the frictional force:
Net Force = Applied Force - Frictional Force
Net Force = 178.0 N * sin(20.0°) - 124.0 N
[Note: We use the sine function because the applied force is at an angle of 20.0° above the horizontal]
Net Force = 178.0 N * 0.3420 - 124.0 N
Net Force = 60.776 N - 124.0 N
Net Force = -63.224 N
[Note: The negative sign indicates that the resulting force is in the opposite direction of the applied force]
2. Using Newton's second law of motion, we can relate the net force to the acceleration of the crate:
Net Force = Mass * Acceleration
-63.224 N = 61.0 kg * Acceleration
Acceleration = -63.224 N / 61.0 kg
Acceleration = -1.036 m/s²
[Note: The negative sign indicates that the crate is decelerating]
3. Now, we can use the kinematic equation to find the final speed of the crate after it has been pulled a distance of 2.90 m. Since the crate starts from rest, its initial speed (u) is 0:
v² = u² + 2 * Acceleration * Distance
v² = 0 + 2 * (-1.036 m/s²) * 2.9 m
v² = -5.997 m²/s²
[Note: The square root of a negative number is not defined in the real number system. However, in this case, we can use the absolute value to find the magnitude of the speed.]
v = sqrt( |-5.997| m²/s²)
v = sqrt(5.997 m²/s²)
v = 2.448 m/s
Therefore, the crate will be moving with a speed of 2.448 m/s after it has been pulled a distance of 2.90 m.