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Ch 04: Newton's Laws of Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 04: Newton's Laws of MotionProblem 10a
Chapter 4, Problem 10a

A dockworker applies a constant horizontal force of 80.080.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.011.0 m in 5.00 5.00 s. What is the mass of the block of ice?

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Start by identifying the known quantities: the applied force \( F = 80.0 \, \text{N} \), the displacement \( d = 11.0 \; \text{m} \), the time \( t = 5.00 \; \text{s} \), and the fact that the block starts from rest (initial velocity \( v_0 = 0 \; \text{m/s} \)). The frictional force is negligible, so the net force is equal to the applied force.
Use the kinematic equation \( d = v_0 t + \frac{1}{2} a t^2 \) to solve for the acceleration \( a \). Since \( v_0 = 0 \), the equation simplifies to \( d = \frac{1}{2} a t^2 \). Rearrange to find \( a \): \( a = \frac{2d}{t^2} \). Substitute \( d = 11.0 \; \text{m} \) and \( t = 5.00 \; \text{s} \) into the equation.
Once the acceleration \( a \) is determined, use Newton's second law \( F = ma \) to solve for the mass \( m \). Rearrange the equation to find \( m \): \( m = \frac{F}{a} \). Substitute \( F = 80.0 \; \text{N} \) and the calculated value of \( a \) into this equation.
Perform the calculations step by step to determine the value of \( a \) and then \( m \). Ensure that the units are consistent throughout the calculations (e.g., \( \text{N} = \text{kg} \cdot \text{m/s}^2 \)).
Finally, verify the result by checking that the calculated mass \( m \) and acceleration \( a \) satisfy both the kinematic equation and Newton's second law. This ensures the solution is consistent and correct.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is expressed by the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, the constant horizontal force applied to the block of ice will cause it to accelerate, allowing us to determine its mass.
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Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. They relate displacement, initial velocity, final velocity, acceleration, and time. In this case, we can use the equation s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time, to find the acceleration of the block of ice and subsequently its mass.
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Acceleration

Acceleration is defined as the rate of change of velocity of an object with respect to time. It can be calculated by dividing the change in velocity by the time taken for that change. In this problem, since the block starts from rest, we can determine its acceleration using the distance it travels and the time taken, which will then help us apply Newton's Second Law to find the mass.
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Related Practice
Textbook Question

A hockey puck with mass 0.1600.160 kg is at rest at the origin (x=0x = 0) on the horizontal, frictionless surface of the rink. At time t=0t = 0 a player applies a force of 0.2500.250 N to the puck, parallel to the xx-axis; she continues to apply this force until t=2.00t = 2.00 s. If the same force is again applied at t=5.00t = 5.00 s, what are the position and speed of the puck at t=7.00t = 7.00 s?

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Textbook Question

A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude 48.048.0 N to the box and produces an acceleration of magnitude 2.202.20 m/s2, what is the mass of the box?

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Textbook Question

You walk into an elevator, step onto a scale, and push the 'up' button. You recall that your normal weight is 625625 N. Draw a free-body diagram. When the elevator has an upward acceleration of magnitude 2.502.50 m/s2, what does the scale read?

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Textbook Question

A 4.504.50-kg experimental cart undergoes an acceleration in a straight line (the xx-axis). The graph in Fig. E4.134.13 shows this acceleration as a function of time. Find the maximum net force on this cart. When does this maximum force occur?

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Textbook Question

You walk into an elevator, step onto a scale, and push the 'up' button. You recall that your normal weight is 625625 N. Draw a free-body diagram. If you hold a 3.853.85-kg package by a light vertical string, what will be the tension in this string when the elevator accelerates as in part (a)? Note: Part (a) asked what does the scale read when the elevator has an upward acceleration of magnitude 2.502.50 m/s2.

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Textbook Question

A hockey puck with mass 0.1600.160 kg is at rest at the origin (x=0x = 0) on the horizontal, frictionless surface of the rink. At time t=0t = 0 a player applies a force of 0.2500.250 N to the puck, parallel to the xx-axis; she continues to apply this force until t=2.00t = 2.00 s. What are the position and speed of the puck at t=2.00t = 2.00 s?

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