Forces and Motion. Basics

4-Wednesday Forces and Motion: Basics Learning Goal: To understand the relationship between applied force, net force, acceleration, and mass for 1-dimensional motion. For this problem, use the PhET simulation Forces and Motion: Basics. This simulation allows you to push on different objects with a certain horizontal force and see the resulting motion. You can control the mass of the objectby using different objects or by stacking up to three objects on lop of one another. Click and drag on the objects to move them to the desired location. You can also adjust Friction between None and Lots by using the slider provided on the right side of the window. While the object is being pushed, you can see the directions and values of the various horizontal forces acting on it, as well as the value of the sum of these forces. In addition you can display the mass, speed, and acceleration of the object. Speed Forces Sum of Forces Values Start the simulation. When you click the simulation link, a new window will open. Double-click on Acceleration to get started. Adjust the Applied Force by moving the slider bar (which is at the center bottom of the main window) left or right. Release the slider to stop applying a force on the object. You can also adjust the Applied Force in increments of 50 Newtons by clicking on the left and right arrows above the slider bar. Apply a force on the crate and watch the resulting motion. Feel free to play around with the simulation, and investigate what happens when you click on either the figure or the object, while they are in motion. When you are done, hit reset (round button with a circular arrow, to the right) and set Friction to None prior to beginning Part A. Make sure that Forces, Sum of Forces, Values, Masses, Speed, and Acceleration are all selected. 13/30 4-Wednesday Choose the 200 m kg refrigerator. Set the applied force to 400 \rm N (to the right). Be sure friction is turned off. What is the net force acting on the refrigerator? ANSWER: The magnitude of the net force is less than 400\rm \;N, directed to the right. The magnitude of the net force is 400\rm \;N, directed to the right. The net force is zero. The magnitude of the net force is greater than 400\rm \;N, directed downward and to the right. The magnitude of the net force is greater than 400\rm \;N, directed upward and to the right. Correct Since there is no friction, the only horizontal force is due to the applied force. The normal force (directed upward) and the force of gravity (directed downward) exactly cancel out. Part B What is the acceleration \texttip{a}{(a) of the refrigerator 4 \m s after the person begins pushing on it with a force of 400 \rm N? Hint 1. How to approach the problem The force acting on an object is F_(net) = ma according to Newton's 2nd law, so if you know the force and the mass of the object, you can solve for the acceleration. ANSWER: (\it a) = 400 \rm N (lit a} = 2\m \;m/s^2 (\it a) = 0.5 \rm \;m/s^2 Correct Applying Newton's 2nd law, (lit a) = F_{net)/m = (400 \rm \;N)/(200 \;kg) = 2 \rm \m/s^2. The acceleration is constant. Part C Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration? Hint 1. How to approach the problem Te refrigerator, then adjust its mass by stacking objects on top of it while it is in motion. Look at the Acceleration indicator as you add different objects. 14/30 4-Wednesday the block itself the earth the surface of the table the string Correct Part K The force acting on the block and directed to the left is ANSWER: a contact force a long-range force Correct Part L What is the force acting on the block and directed to the left called? ANSWER: tension normal force weight friction Correct Now consider a slightly different situation. The same block is placed on the same rough table. However, this time, the string is disconnected and the block is given a quick push to the right. The block slides to the right and eventually stops. The following questions refer to the motion of the block affer it is pushed but before it stops. Part M How many forces are acting on the block in the horizontal direction? ANSWER: 23/30 4-Wednesday Correct 00 1 2 3 Once the push has ended, there is no force acting to the right: The block is moving to the right because it was given a velocity in this direction by some force that is no longer applied to the block (probably, the normal force exerted by a student's hand or some spring launcher). Once the contact with the launching object has been lost, the only horizontal force acting on the block is directed to the left--which is why the block eventually stops. Part N What is the force acting on the block that is directed to the left called? ANSWER: tension normal force weight friction Correct The force of friction does not disappear as long as the block is moving. Once the block stops, fricion becomes zero (assuming the table is perfectly horizontal). Conceptual Questions on Newton's 1st and 2nd Laws Learning Goal: To understand the meaning and the basic applications of Newton's 1st and 2nd laws. In this problem, you are given a diagram representing the motion of an object--a motion diagram. The dots represent the object's position at moments separated by equal intervals of time. The dots are connected by arrows representing the object's average velocity during the corresponding time interval. Your goal is to use this motion diagram to determine the direction of the net force acting on the object. You will then determine which force diagrams and which situations may correspond to such a motion. 24/30 4-Wednesday Part A What is the direction of the net force acting on the object at position A? Hint 1. Using Newton's 2nd law According to Newton's 2nd law, vectors \texttip(\vec{a}}{a_vec) and \vec{F}_{\rm net) have the same direction. Can you determine the direction of acceleration at position A by analyzing the diagram? ANSWER: Bupward downward to the left to the right The net force is zero. Correct The velocity vectors connecting position A to the adjacent positions appear to have the same magnitude and direction. Therefore, the acceleration is zero-and so is the net force. Part B What is the direction of the net force acting on the object at position B? ANSWER: 25/30