A constant horizontal force of 275 N was applied to push a crate of mass
m = 92.0 kg at a speed of v = 0.850 m/s, which encounters a rough horizontal
surface of length, l = 0.65 m as in Figure 2 below. If the coefficient of kinetic
friction between the crate and rough surface is 0.358, find
(a) The magnitude and direction of the net force on the crate while it is on
the rough surface,
(b) The net work done on the crate while it is on the rough surface, and
(c) The speed of the crate when it reaches the end of the rough surface.
To find the magnitude and direction of the net force on the crate while it is on the rough surface, we need to consider the forces acting on the crate.
The forces acting on the crate are the applied force, the normal force, and the frictional force.
(a) The normal force is the force exerted by the surface perpendicular to the crate. On a horizontal surface, the normal force is equal in magnitude and opposite in direction to the weight of the object. So, the normal force is given by:
Normal force = mg,
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Normal force = (92.0 kg)(9.8 m/s^2) = 901.6 N.
The frictional force can be found using the equation:
Frictional force = coefficient of kinetic friction × normal force.
Frictional force = (0.358)(901.6 N) = 322.5 N.
Since the crate is moving in the positive direction of the applied force, the frictional force will act in the opposite direction (negative direction). Therefore, the magnitude and direction of the net force on the crate while it is on the rough surface is 322.5 N in the negative direction.
(b) The work done on an object is given by the equation:
Work done = force × distance × cos(theta),
where force is the net force on the object, distance is the displacement of the object, and theta is the angle between the force and displacement vectors.
Since the crate is moving in the positive direction of the applied force, the angle between the force and displacement vectors is 0 degrees. Therefore, cos(theta) = 1.
The distance the crate moves on the rough surface is given as l = 0.65 m.
Work done = (net force) × (distance) × (cos(theta)) = (322.5 N) × (0.65 m) × (1) = 209.6 J.
So, the net work done on the crate while it is on the rough surface is 209.6 J.
(c) The speed of the crate when it reaches the end of the rough surface can be found using the work-energy theorem. The work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy of the crate is given by:
Initial kinetic energy = (1/2) × (mass) × (initial velocity)^2.
The final kinetic energy of the crate is given by:
Final kinetic energy = (1/2) × (mass) × (final velocity)^2.
Since the crate started from rest at the beginning of the rough surface, the initial kinetic energy is zero.
The work done on the crate is equal to the change in kinetic energy:
Work done = Final kinetic energy - Initial kinetic energy.
209.6 J = (1/2) × (92.0 kg) × (final velocity)^2 - 0.
Rearranging the equation:
(final velocity)^2 = (2 * Work done) / (mass).
(final velocity)^2 = (2 * 209.6 J) / (92.0 kg).
(final velocity)^2 = 4.557 m^2/s^2.
Taking the square root of both sides:
final velocity = sqrt(4.557 m^2/s^2) = 2.135 m/s.
So, the speed of the crate when it reaches the end of the rough surface is 2.135 m/s.