A rope exerts a constant horizontal force of 203N pulling a 68-kg crate across the floor. The velocity of the crate is observed to increase from 3m/s to 5m/s in a time of 4 seconds under the influence of this force and the frictional force exerted by the floor on the crate.

see other post

To analyze the given information about the rope pulling the crate, we can consider Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).

In this case, the horizontal force exerted by the rope is 203N, and the mass of the crate is 68kg. We can assume that the frictional force exerted by the floor on the crate is in the opposite direction and is the cause of the crate's acceleration.

Since the crate's velocity changes from 3m/s to 5m/s in a time of 4 seconds, we can find the acceleration using the formula:

Acceleration (a) = (final velocity - initial velocity) / time taken
a = (5 m/s - 3 m/s) / 4 s
a = 2 m/s^2

Now that we have the acceleration, we can calculate the net force:

Net force (F_net) = mass * acceleration
F_net = 68 kg * 2 m/s^2
F_net = 136 N

The net force includes both the force exerted by the rope and the force opposing the motion due to friction. So, we can calculate the frictional force:

Frictional force (F_friction) = F_net - Force exerted by the rope
F_friction = 136 N - 203 N
F_friction = -67 N

The negative sign indicates that the frictional force is in the opposite direction to the force exerted by the rope.

Hence, the frictional force exerted by the floor on the crate is 67N.