Question 1 0/1A 100 kg person stands with their left foot on one bathroom scale, and their right foot on a second bathroom scale. Both scales display weight in Newtons. If their weight is divided evenly between each foot, how many Newtons does the left bathroom scale display?Your Answer: 4,900 Question 2 0/1A 100 kg person stands with their left foot on one bathroom scale, and their right foot on a second bathroom scale. Both scales display weight in Newtons. If they have 60% of their weight on their left foot and the remaining weight on their right foot, how many Newtons does the right bathroom scale display?Your Answer: 5,880 Question 3 0/1A 100 kg person stands with their left foot on one bathroom scale, and their right foot on a second bathroom scale. Both scales display weight in Newtons. If the left scale reads twice as much as the right scale, what does the right scale display?Your Answer: 3,267 Question 4 0/1A window washer weighing 800 N stands on a stationary scaffold supported by two ropes. The left rope has 600 N of tension, and the right rope has 400 N of tension. What is the weight of the scaffold in Newtons? Question 17A 100 kg person rides in an elevator moving at a constant speed of 3 m/s downward.What is the normal force (in N) on the person’s feet?Your Answer: 0 Question 18 0/1Two people play tug of war. The 110 kg person on the left pulls with 1000 N of force, and the 60 kg person on the right pulls with 700 N of force.What is the acceleration (in m/s2) of the right-side person?Your Answer: 1.76 Question 19 0/1Two people play tug of war. The 110 kg person on the left pulls with 1000 N of force, and the 60 kg person on the right pulls with 700 N of force.What is the acceleration (in m/s2) of the left-side person?Your Answer: 1.76 Question 20 0/1Two people play tug of war. The 110 kg person on the left pulls with 1000 N of force, and the 60 kg person on the right pulls with 700 N of force.What is the tension in the rope (in N) pulling on the left side person?Your Answer: Question 21 0/1A hanging mass, m = 5 kg, hangs from a string that goes over a pulley and pulls on a mass, M = 10 kg, that sits on a table as shown below. The coefficient of friction between the sitting mass and the surface is 0.3.What is the magnitude (in m/s2) of the acceleration of the hanging mass?Your Answer: Question 22 0/1A hanging mass, m = 5 kg, hangs from a string that goes over a pulley and pulls on a mass, M = 10 kg, that sits on a table as shown below. The coefficient of friction between the sitting mass and the surface is 0.3.What is the magnitude of the tension (in N) in the rope holding the hanging mass?Your Answer: Question 23 0/1100 kg person rides in an elevator accelerating 2 m/s2 upward. (Hint: Watch “Hewitt-Drew-it! PHYSICS 23: Nellie in an Elevator” in which he solves an almost identical problem. This is just a net force problem like the rest, so use your net force equation.) What is the normal force (in N) acting on the person’s feet?Your Answer: Question 24 0/1100 kg person rides in an elevator accelerating 2 m/s2 upward.How does the magnitude of the normal force exerted by the floor on the person’s feet compare to the magnitude of their weight (in N)?Show answer choices Normal force magnitude is less than weight magnitudeNormal force magnitude is equal to weight magnitudeNormal force magnitude is greater than weight magnitudeNot enough information to determine Question 25 0/1100 kg person rides in an elevator accelerating 2 m/s2 upward.What value (in m/s2) would the acceleration have to be for the magnitude of the normal force exerted by the floor on the person’s feet to be equal to the magnitude of their weight? (Use positive values for an upwards acceleration, and use negative values for a downward acceleration.) Extensions Derive one equation that relates the horizontal and vertical coordinates of the balls motion in this experiment (gives y as a function of x, y(x)). Hint: Solve the x equation of motion for time, then substitute into the y equation of motion to obtain an equation that gives y in terms of x. Derive a general formula y(x) for projectile motion with the object launched at an angle at a given velocity . The resulting equation should be y in terms of the variable x and the two parameters and . Hint: You can use the result of the previous problem to derive this expression quickly. Calibrate the velocity of the ball when released from various positions along the ramp. Given a specific distance to the target by the instructor, determine where the ball must be released to achieve the needed velocity. Release the ball from that position and determine whether the target is hit.
