week 3 assignment - PHYS133

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University of Hawaii, Hilo *

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Physics

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Dec 6, 2023

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Mariah Keller Castillo 5786597 PHYS133 K002

Chapter 4: problem 3 To calculate the magnitude of the acceleration of the laundry cart, you can use Newton's second law of motion, which states: F=maF=ma Where:  FF is the net external force applied to the object (in this case, 60.0 N).  mm is the mass of the object (4.50 kg).  aa is the acceleration of the object (what we want to find). Now, you can rearrange the formula to solve for acceleration (aa): a=Fma=mF Plug in the values: a=60.0 N4.50 kga=4.50kg60.0N Now, calculate the acceleration: a=60.0 N4.50 kg≈13.33 m/s2a=4.50kg60.0N≈13.33m/s2 So, the magnitude of the acceleration of the laundry cart is approximately 13.33 m/s213.33m/s2. Chapter 4: problem 16 To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. We'll consider each part of the problem separately. (a) What is the force of friction between the losing player's feet and the grass? In this case, we have a losing player who is being pushed backward, so there must be a force of friction acting in the opposite direction to the motion. We can calculate this force of friction using the following equation: Ffriction=m ⋅ aFfriction=m ⋅ a Where: FfrictionFfriction = Force of friction mm = Mass of the losing player plus equipment = 90.0 kg aa = Acceleration of the losing player = 2.20 m/s² (negative because it's in the opposite direction)

Ffriction=90.0kg ⋅ (−2.20m/s2)=−198NFfriction=90.0kg ⋅ (−2.20m/s2)=−198N The force of friction between the losing player's feet and the grass is 198 N, and it acts in the opposite direction to the motion. (b) What force does the winning player exert on the ground to move forward? In this case, the winning player is exerting a force on the losing player, causing both players to move in opposite directions. To find the force exerted by the winning player on the ground, we can use Newton's third law, which states that action and reaction are equal and opposite. So, the force exerted by the winning player on the ground is equal in magnitude but opposite in direction to the force exerted on the losing player. Therefore, it is also 800 N but in the opposite direction (forward). (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free body diagram and write the net force equation. I can describe how to draw a sketch and a free body diagram: Sketch: Draw two rugby players, one labeled as the losing player and the other as the winning player.  Show an arrow pointing to the right (opposite to the direction of motion) with a label of 800 N for the force applied by the winning player on the losing player.  Show another arrow pointing to the left (the direction of motion) with a label of 198 N for the force of friction acting on the losing player.  Label the mass of the losing player as 90.0 kg and the mass of the winning player as 110 kg. Free Body Diagram for the Losing Player:  Draw a dot to represent the losing player.  Draw an arrow pointing to the right to represent the force applied by the winning player (800 N).  Draw an arrow pointing to the left to represent the force of friction (198 N).  Label the mass of the losing player as 90.0 kg.

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Net Force Equation for the Losing Player: Fnet=Fapplied+FfrictionFnet=Fapplied+Ffriction Fnet=800N−198NFnet=800N−198N Fnet=602NFnet=602N So, the net force on the losing player is 602 N in the direction of motion (left). This force causes the losing player to accelerate backward at 2.20 m/s². Chapter 4: problem 20 To find the tension in the rope, we'll use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the gymnast is climbing a rope, so we need to consider two scenarios: one where the gymnast climbs at a constant speed (zero acceleration), and the other where the gymnast accelerates upward at a rate of 1.50 m/s². (a) Tension in the rope when the gymnast climbs at a constant speed (zero acceleration): When the gymnast climbs at a constant speed, the net force on the gymnast is zero because there is no acceleration. Therefore, the tension in the rope must be equal in magnitude to the gravitational force acting on the gymnast. The gravitational force can be calculated using the formula: Fgravity=m ⋅ gFgravity=m ⋅ g Where: FgravityFgravity = Gravitational force mm = Mass of the gymnast = 60.0 kg gg = Acceleration due to gravity ≈ 9.81 m/s² Fgravity=60.0kg ⋅ 9.81m/s2=588.6NFgravity=60.0kg ⋅ 9.81m/s2=588.6N So, when the gymnast climbs at a constant speed, the tension in the rope is 588.6 N, acting upward. (b) Tension in the rope when the gymnast accelerates upward at a rate of 1.50 m/s²: In this case, the gymnast is accelerating upward, so we need to consider the net force required to produce this acceleration. The net force is given by: Fnet=m ⋅ aFnet=m ⋅ a Where: FnetFnet = Net force (tension in the rope) mm = Mass of the gymnast = 60.0 kg aa = Acceleration = 1.50 m/s² (upward)

Fnet=60.0kg ⋅ 1.50m/s2=90.0NFnet=60.0kg ⋅ 1.50m/s2=90.0N So, when the gymnast accelerates upward at a rate of 1.50 m/s², the tension in the rope is 90.0 N, acting upward. To summarize: (a) Tension in the rope when the gymnast climbs at a constant speed is 588.6 N (upward). (b) Tension in the rope when the gymnast accelerates upward at a rate of 1.50 m/s² is 90.0 N (upward). Chapter 4: problem 30 (a) Finding the Magnitude of Forces F1 and F2: You can find the magnitude of forces F1 and F2 by using trigonometry if you know the angle between them and the magnitude of the total force Ftot. Let θ be the angle between F1 and Ftot (or F2 and Ftot). You can use the following trigonometric relationship: F1=Ftot ⋅ cos(θ)F1=Ftot ⋅ cos(θ) F2=Ftot ⋅ sin(θ)F2=Ftot ⋅ sin(θ) This allows you to find the magnitudes of F1 and F2. (b) Showing the Same Total Force Independent of Order: To show that the same total force Ftot is obtained regardless of the order of addition, you can use the triangle law of vector addition. Suppose you have F1 and F2, and you want to find Ftot.  Draw vector F1 from the origin to its tip.  From the tip of F1, draw vector F2. Now, the total force Ftot is the vector that goes from the origin to the tip of F2. It doesn't matter if you draw F1 first and then F2 or vice versa; you'll end up at the same point, representing the same total force Ftot. (c) Finding Another Pair of Vectors: To find another pair of vectors that add up to give Ftot, you can use the parallelogram law of vector addition. Start with F1 and F2 as given. Draw F1 and F2 as adjacent sides of a parallelogram. The diagonal of the parallelogram, which starts at the same point as F1 and ends at the same point as F2, represents Ftot.

So, you can find another pair of vectors that add up to Ftot by drawing any pair of vectors that can be represented as the sides of the same parallelogram.

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