A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point A , which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray? ( Hint: This does not occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point A and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?
A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point A , which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray? ( Hint: This does not occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point A and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?
A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point A, which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point A and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?
The force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE DO NOT USE LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!
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The force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE ONLY TRIGNOMETRIC FUNCTIONS (SIN/TAN/COS, NO LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!
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