3. A 2.60 kg mass is being pulled by a force of 19.6 N at an angle of elevation of 35.0° as shown in the diagram below. The coefficient of friction between the floor and the block is 0.270. If the block starts from rest, what is its speed after being pulled with this force for 11.0 s? Hint: find the net acceleration first. (23 marks)

4. In order to solve the acceleration question, one must follow these steps: 1.Draw a free-body diagram. 2.Find the x- and y-components of the applied force vector. 3.Find the net force in the y-direction by adding the applied force in the y- direction to the weight if it’s being pushed and subtract the applied force in the y-direction if it’s being pulled. 4.Use the net force in the y-direction to calculate the force of friction. 5.Calculate the net force in the x-direction by subtracting the frictional force from the applied force in the x-direction. 6.Calculate the net acceleration by dividing the net force in the x-direction by the mass of the block. 7.Calculate final velocity using kinematics equations.

5. FgFg F app F appx FfFf To understand what is happening in this problem, one must first construct a free-body diagram to describe the forces involved. F appy F f is the force of friction acting on the mass. F app is the applied force acting on the mass. F appy is the component of the applied force acting in the y-direction. F appx is the component of the applied force acting in the x-direction. F g is the force of gravity (weight) acting on the mass.

6. In order to solve for the final speed of the block, we must first break down the applied force into its x- and y- components. F app F appy Because this block is being pulled, the force acting in the y-direction is an upward force as seen here. If the block were being pushed, the force in the y- direction would be a downward force and it would add to the weight of the object being pushed. F app F appy

7. There is a component of the applied force that acts in the direction of motion, or the x-direction. We calculate this force by using the cos function. F app F appx This component in the x-direction will be opposed by the force of friction. However, what must first be found is the net force acting in the y-direction.

8. It is necessary for us to calculate the effective weight (net force acting in the y-direction) of the block, given the fact that it has its own weight MINUS the component of the pulling force that we have applied. The effective weight is increased when we are pushing because we are adding to the weight, but it is decreased when we are pulling the mass because we out y-component applies a lifting force to the mass. FgFg F app F appx FfFf F appy

9. It is necessary for us to calculate the effective weight (net force acting in the y-direction) of the block, given the fact that it has its own weight MINUS the component of the pulling force that we have applied. The effective weight is increased when we are pushing because we are adding to the weight, but it is decreased when we are pulling the mass because we out y-component applies a lifting force to the mass. FgFg F appy F nety

10. Once you have calculated the net force in the y-direction, you can then calculate the net force acting in the x-direction because the net force in the y-direction is needed to calculate the force of friction. One must finally calculate the net acceleration of the block using Newton’s 2 nd Law. F app F appx FfFf F appy

11. Once you have calculated the net force in the y-direction, you can then calculate the net force acting in the x-direction because the net force in the y-direction is needed to calculate the force of friction. One must finally calculate the net acceleration of the block using Newton’s 2 nd Law. F appx FfFf F netx

12. At this point the question merely becomes a kinematic question. We solve for the final velocity beginning with the definition of acceleration. F app