Sally applies a horizontal force of 500 N with a rope to dtag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 67.5 degrees. a) How much force is exerted by the rope on the crate? b) What work is done by Sally if the crate is moved 30.0 m? c) What work is done by the floor through force of friction between the floor and the crate?

Here is what I know: Work = Force x Distance and being as there is not a change in velocity there is not a net force. Therefore, Force applied = Force of friction. Once work is found the other two questions should be easy to answer.

The 500 N Sally applies must be the horizontal component of the appied force, not the total force (which is the ripe tension)

The horizontal force component is
T cos 67.5 = 500 N
Therefore T = 1306 N
b) Sally's Work = (horizontal force component) x (distance moved)
= 500 N x 30 m = ___ J
c) -(Work done by Sally)

Thank you!!

a) To find the force exerted by the rope on the crate, we need to resolve the applied force into its horizontal and vertical components.

The horizontal component (Fx) can be found using the formula: Fx = Force * cos(angle)

Fx = 500 N * cos(67.5 degrees) ≈ 500 N * 0.386 ≈ 193 N

b) The work done by Sally can be calculated using the formula: Work = Force * Distance

Work = Fx * Distance

Work = 193 N * 30.0 m = 5,790 N*m (or Joules)

c) Since the crate is moving at a constant speed, we know that the force of friction is equal in magnitude but opposite in direction to the applied force. Hence, the work done by the floor through the force of friction is:

Work = Force of friction * Distance

Work = Fx * Distance

Work = 193 N * 30.0 m = 5,790 N*m (or Joules)

So, the work done by the floor through the force of friction is also 5,790 N*m (or Joules).

To find the force exerted by the rope on the crate, we need to decompose the applied force into its horizontal and vertical components.

The horizontal component of the force can be found using the equation Fcosθ, where F is the magnitude of the force (500 N) and θ is the angle between the force and the horizontal direction (67.5 degrees).

So, the horizontal component of the force exerted by the rope on the crate is 500 N * cos(67.5 degrees).

a) To find the force exerted by the rope on the crate, simply calculate the horizontal component using the provided formula:
Force exerted by the rope on the crate = 500 N * cos(67.5 degrees).

Now, let's move on to the next question.

b) To find the work done by Sally, we can use the formula: Work = Force x Distance.

Here, the force exerted by Sally is the horizontal component of the force, which we calculated in part a), and the distance is given as 30.0 m. Plugging in the values, we get:
Work done by Sally = (500 N * cos(67.5 degrees)) * 30.0 m.

Finally, let's move to the last question.

c) Since the crate is moving with a constant speed, there is no change in its kinetic energy. This means that the work done by the floor through the force of friction is equal in magnitude but opposite in sign to the work done by Sally.

Therefore, the work done by the floor through the force of friction can be calculated as:
Work done by the floor = -(Work done by Sally).

To sum up, a) the force exerted by the rope on the crate is 500 N * cos(67.5 degrees), b) the work done by Sally is (500 N * cos(67.5 degrees)) * 30.0 m, and c) the work done by the floor through the force of friction is -(Work done by Sally).