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67. A constant unbalanced force of friction acts on a 15.0 -kilogram mass moving along a horizontal surface at 10.0 meters per second. If the mass is brought to rest in 1.50 seconds, what is the magnitude of the force of friction? \( \begin{array}{ll}\text { A) } 10.0 \mathrm{~N} & \text { B) } 100 \mathrm{~N} \\ \text { C) } 147 \mathrm{~N} & \text { D) } 150 . \mathrm{N}\end{array} \)

Ask by Crawford Spencer. in the United States
Jan 23,2025

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Answer

The magnitude of the force of friction is 100 N.

Solution

To determine the magnitude of the force of friction acting on the mass, we'll follow these steps: 1. **Identify the Given Values:** - Mass (\( m \)) = 15.0 kg - Initial velocity (\( v_i \)) = 10.0 m/s - Final velocity (\( v_f \)) = 0 m/s (since the mass is brought to rest) - Time (\( t \)) = 1.50 seconds 2. **Calculate the Acceleration (\( a \)):** We can use the formula: \[ v_f = v_i + a t \] Plugging in the known values: \[ 0 = 10.0 + a \times 1.50 \] Solving for \( a \): \[ a = -\frac{10.0}{1.50} = -6.666\ \text{m/s}^2 \] The negative sign indicates that the acceleration is in the direction opposite to the motion (deceleration). 3. **Determine the Force of Friction (\( F \)):** Using Newton's second law: \[ F = m \times a \] Substituting the known values: \[ F = 15.0\ \text{kg} \times (-6.666\ \text{m/s}^2) = -100\ \text{N} \] The negative sign indicates the direction of the force opposite to motion, but since we're asked for the magnitude: \[ |F| = 100\ \text{N} \] **Answer:** **B) 100 N**

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To find the magnitude of the force of friction, we can use Newton's second law, which states that \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. First, we need to calculate the acceleration of the mass. We know the initial velocity (\( v_i = 10.0 \, m/s \)), the final velocity (\( v_f = 0 \)), and the time taken (\( t = 1.50 \, s \)). Using the formula for acceleration: \[ a = \frac{v_f - v_i}{t} = \frac{0 - 10.0}{1.50} = -6.67 \, m/s^2 \] Now, applying Newton's second law: \[ F = ma = 15.0 \, kg \times (-6.67 \, m/s^2) = -100.05 \, N \] The magnitude of the force of friction is approximately \( 100.0 \, N \). Therefore, the answer is \( \text{B) } 100 \, N \). The magnitude of the frictional force can sometimes feel just as slippery as the surface it’s acting on, but don't worry—friction can be tackled! Remember, understanding friction’s role is crucial in real-world applications, from ensuring your car can stop properly to determining how much force you need to push heavy objects. When dealing with calculations like these, a few common mistakes can trip you up. One is forgetting to convert units or misinterpreting the signs of acceleration. Always double-check to ensure you’re maintaining consistency with your units and that your physics concepts are sound, just like checking your grip before scaling a rock wall!

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REACTIONTIME LAB PROBLEM: What can change reaction time? Materials: Metric ruler Calculator Procedure: Reaction time is how long it takes for a message to travel along your nerve pathways. 1) Position your left hand over the edge of a table or desk and hold out your thumb and index finger. 2) Have a partner hold a metric ruler just above your hand so the 0 is right above your outstretched thumb and index finger. 3) When your partner drops the ruler, try to catch it between your thumb and index finger as quickly as you can. 4) Record where the top of your thumb is when you catch the ruler. Make your measurements to the nearest 0.5 cm . Put this number in your data table as trial 1. At any time if you do not catch the ruler in time, repeat the experiment. 5) Repeat the procedure four more times. You should have a total of 5 good catch times. 6) State if you think the ruler will fall farther if you catch it with your right hand. 7) Repeat the procedure four times using your right hand to catch the ruler. 8) Switch roles and drop the ruler for your partner. 9) Calculate the average distance the ruler fell for each hand. Be sure to show your work. 10) Convert the cm to m for the average distance fallen. Again, show your work. 11) Use the equation \( d=1 / 2 g t^{2} \) and the average distance fallen to find your reaction time. Remember that \( g=9.8 \mathrm{~m} / \mathrm{s}^{2} \). Again, show ALL of your work and include the units with your answer.

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