A 50.0kg box is sliding to the right and coming to a stop. The coefficient of kineticfriction is: 0.40a. Draw a free body diagram (2 marks)b. What is the net force? (2 marks)c. What is the net acceleration? (2 marks)d. What was the initial speed if it took 3.0s to stop? (2 marks)e. How much displacement was covered before stopping? (2 marks)

a. A free body diagram would include the following forces:

• The force of gravity (Fg) acting downwards, which is equal to the mass of the box times the acceleration due to gravity (50.0 kg * 9.8 m/s^2 = 490 N).
• The normal force (Fn) acting upwards, which is equal to the force of gravity (490 N) since the box is not accelerating vertically.
• The force of friction (Ff) acting to the left, which is equal to the coefficient of kinetic friction times the normal force (0.40 * 490 N = 196 N).

b. The net force is the force of friction, which is acting to the left (or in the negative direction). So, the net force is -196 N.

c. The net acceleration can be found using Newton's second law (F = ma). Rearranging for acceleration gives a = F/m. Substituting the given values gives a = -196 N / 50.0 kg = -3.92 m/s^2. The negative sign indicates that the acceleration is to the left, which makes sense since the box is slowing down.

d. The initial speed can be found using the equation v = u + at, where v is the final speed, u is the initial speed, a is the acceleration, and t is the time. Since the box comes to a stop, the final speed is 0. Rearranging the equation for initial speed gives u = v - at = 0 - (-3.92 m/s^2 * 3.0 s) = 11.76 m/s.

e. The displacement can be found using the equation s = ut + 0.5at^2, where s is the displacement, u is the initial speed, a is the acceleration, and t is the time. Substituting the given values gives s = 11.76 m/s * 3.0 s + 0.5 * -3.92 m/s^2 * (3.0 s)^2 = 35.28 m - 17.64 m = 17.64 m.