A worker applies a constant force of magnitude F = 339 N on a crate in a direction è = 26.1o above the horizontal as he pulls it across a level floor through a distance of s = 7.20 m. The mass of the crate is m = 53 kg and the coefficient of kinetic friction between the floor and the crate is ìk = 0.255.

1) Calculate the amount of work done on the crate by the worker
2) Calculate the amount of work done on the crate by friction
3) Calculate the net force acting on the crate in the x-direction
4) Use the net force to calculate the total work done on the crate
5) Assuming the crate starts from rest, calculate the final velocity of the crate

my problem is finding number 1 and 2 and 3.. please explain how to do this for me! thank you!

btw i found that both the amount of work done on the crate by normal force and gravity are both 0.

Sure! I can help you with those questions. Let's go step by step.

1) To calculate the amount of work done on the crate by the worker, you can use the formula:

Work = Force x Distance x cos(θ)

In this case, the force applied by the worker is F = 339 N, and the distance over which the force is applied is s = 7.20 m. The angle between the force and the direction of motion is θ = 26.1°. Plugging in these values into the formula, we get:

Work = 339 N x 7.20 m x cos(26.1°)

Calculating this will give you the answer to the first question.

2) To calculate the amount of work done on the crate by friction, you can use the formula:

Work = Force of friction x Distance

The force of friction can be found using the equation:

Force of friction = μk x Normal force

Given that the coefficient of kinetic friction is μk = 0.255 and the normal force is equal to the weight of the crate (mg), where m is the mass of the crate (53 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now, you can calculate the force of friction and then use it in the work formula along with the distance s = 7.20 m to find the work done on the crate by friction.

3) To find the net force acting on the crate in the x-direction, you need to consider the horizontal components of all forces acting on the crate. The vertical forces (gravity, normal force) do not contribute to the motion in the x-direction.

The horizontal component of the applied force is given by Fx = Fcos(θ), where F is the magnitude of the force and θ is the angle with the horizontal direction.

The force of friction acts in the opposite direction, so its horizontal component is -Ff, where Ff is the force of friction. Therefore, the net force in the x-direction is:

Net force (in x-direction) = Fx - Ff

Calculate this net force using the known values.

I hope this helps! If you have any other questions, feel free to ask.