/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A dock worker applies a constant... [FREE SOLUTION] | 91Ó°ÊÓ

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A dock worker applies a constant horizontal force of 80.0 \(\mathrm{N}\) to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 \(\mathrm{m}\) in the first 5.00 \(\mathrm{s}\) s. What is the mass of the block of ice?

Short Answer

Expert verified
The mass of the block of ice is approximately 90.91 kg.

Step by step solution

01

Identify the Known Values

The force applied, \( F = 80.0 \, \mathrm{N} \). The block moves a distance of \( s = 11.0 \, \mathrm{m} \) in time \( t = 5.00 \, \mathrm{s} \). Since the block starts from rest, the initial velocity \( u = 0 \, \mathrm{m/s} \).
02

Use Kinematic Equation to Find Acceleration

The kinematic equation \( s = ut + \frac{1}{2}a t^2 \) is used to find the acceleration \( a \). Substituting the values: \( 11.0 = 0 + \frac{1}{2}a (5.00)^2 \). Simplifying, we have \( 11.0 = \frac{1}{2} a \times 25 \), so \( a = \frac{22}{25} = 0.88 \mathrm{m/s^2} \).
03

Use Newton's Second Law to Find Mass

According to Newton's second law, \( F = ma \). Rearrange this equation to solve for mass \( m \): \( m = \frac{F}{a} \). Substitute the known values: \( m = \frac{80.0 \, \mathrm{N}}{0.88 \, \mathrm{m/s^2}} \). Calculating gives \( m \approx 90.91 \, \mathrm{kg} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations
Kinematic equations are a crucial part of physics, especially in problems involving motion. They help in describing the motion of objects, providing a relationship between variables like displacement, velocity, acceleration, and time. In this exercise, the kinematic equation used is:
  • \( s = ut + \frac{1}{2} a t^2 \)
This equation relates the displacement \( s \) of an object, its initial velocity \( u \), the acceleration \( a \), and the time \( t \). Since the block of ice starts from rest, the initial velocity \( u = 0 \). By plugging in the known values, we calculate the acceleration \( a \) of the block. The equation simplifies to \( 11.0 = \frac{1}{2} a (5.00)^2 \), which gives us \( a = 0.88 \, \mathrm{m/s^2} \). Understanding how to manipulate these equations allows for solving complex motion problems.
Constant Force
Applying a constant force to an object leads to a uniform change in motion, provided there is no opposing force like friction. In this exercise, a constant force of 80.0 \( \mathrm{N} \) is applied to the block of ice. Newton's Laws tell us how this constant force affects the motion.
  • Since this force is constant and horizontal, and friction is negligible, all of the force contributes to moving the block.
  • The constant force ensures a constant acceleration, assuming mass stays the same.
Thus, the block of ice experiences a steady increase in speed, driven purely by the applied force.
Acceleration Calculation
Acceleration is the rate of change of velocity and is a critical factor in determining how quickly an object speeds up or slows down. To find acceleration in this scenario, we used the kinematic equation:
  • \( s = ut + \frac{1}{2} a t^2 \)
We substituted the values to get \( a = 0.88 \, \mathrm{m/s^2} \). This means that every second, the velocity of the block increases by 0.88 meters per second. Acceleration here is constant because the force applied is consistent and there is no friction to alter the motion. Calculating acceleration accurately is essential as it affects the final calculation of mass through Newton's Second Law.
Mass Determination
Mass determination involves understanding the relationship between force, mass, and acceleration, as expressed in Newton's Second Law of Motion:
  • \( F = ma \)
Here, \( F \) is the force applied, \( m \) is the mass, and \( a \) is acceleration. We rearrange the equation to find the mass:
  • \( m = \frac{F}{a} \)
By substituting the known values into the equation, \( m = \frac{80.0 \, \mathrm{N}}{0.88 \, \mathrm{m/s^2}} \), we calculate that the mass of the block is approximately \( 90.91 \, \mathrm{kg} \). Mass is a fundamental property, and accurately determining it allows us to understand how much matter is present in the block of ice, influencing how it responds to the applied force.

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Most popular questions from this chapter

(a) How many newtons does a 150 lb person weigh? (b) Should a veterinarian be skeptical if someone said that her adult collie weighed 40 \(\mathrm{N} ?(\mathrm{c})\) Should a nurse question a medical chart which showed that an average-looking patient had a mass of 200 \(\mathrm{kg} ?\)

A standing vertical jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 \(\mathrm{m}\) (4 ft). (This means that he moved upward by 1.2 \(\mathrm{m}\) after his feet left the floor.) Griffith weighed 890 \(\mathrm{N}(200 \mathrm{lb}) .\) (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what were the magnitude and direction of his acceleration (assuming it to be constant) while he was pushing against the floor? (c) Draw a free-body diagram of Griffith during the jump. (d) Use Newton's laws and the results of part (b) to calculate the force he applied to the ground during his jump.

Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is \(60.0^{\circ} .\) If \(\operatorname{dog} A\) exerts a force of 270 \(\mathrm{N}\) and dog \(B\) exerts a force of \(300 \mathrm{N},\) find the magnitude of the resultant force and the angle it makes with dog \(A\) 's rope.

A rifle shoots a 4.20 g bullet out of its barrel. The bullet has a muzzle velocity of 965 \(\mathrm{m} / \mathrm{s}\) just as it leaves the barrel. Assuming a constant horizontal acceleration over a distance of 45.0 \(\mathrm{cm}\) starting from rest, with no friction between the bullet and the barrel, (a) what force does the rifle exert on the bullet while it is in the barrel? (b) Draw a free-body diagram of the bullet (i) while it is in the barrel and (ii) just after it has left the barrel. (c) How many \(g^{\prime}\) s of acceleration does the rifle give this bullet? (d) For how long a time is the bullet in the barrel?

A woman is standing in an elevator holding her 2.5 -kg briefcase by its handles. Draw a free-body diagram for the briefcase if the elevator is accelerating downward at 1.50 \(\mathrm{m} / \mathrm{s}^{2}\) and calculate the downward pull of the briefcase on the woman's arm while the elevator is accelerating.

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