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ssm A dump truck is being filled with sand. The sand falls straight downward from rest from a height of 2.00 m above the truck bed, and the mass of sand that hits the truck per second is 55.0 kg/s. The truck is parked on the platform of a weight scale. By how much does the scale reading exceed the weight of the truck and sand?

Short Answer

Expert verified
The scale reading exceeds the weight by 484 N.

Step by step solution

01

Understand the Problem

We need to determine how much the scale reading increases when sand falls onto the truck bed at a rate of 55.0 kg/s from a height of 2.00 m. This involves calculating the additional force exerted by the falling sand.
02

Calculate the Speed of the Sand Upon Impact

Use the kinematic equation to find the velocity of the sand just before it hits the truck bed. The equation is \[ v = \sqrt{2gh} \]where \( g = 9.8 \text{ m/s}^2 \) (acceleration due to gravity) and \( h = 2.00 \text{ m} \). Plug in these values to find \( v \).
03

Determine the Momentum Change Per Second

The change in momentum (also known as the rate of momentum change) is equal to the force exerted by the falling sand. This can be found using \[ F = \frac{d}{dt}(mv) = mv \]where \( m = 55.0 \text{ kg/s} \) is the mass flow rate, and \( v \) is the velocity calculated in Step 2.
04

Calculate the Additional Force on the Scale

Multiply the mass flow rate by the velocity to get the additional force:\[ F = 55.0 \text{ kg/s} \times v \]
05

Find the Increase in Scale Reading

The increase in force due to the sand's momentum change represents the increase in the scale reading over the weight of the truck and static sand. Thus, the additional weight reading is \( F \), found in Step 4.

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Key Concepts

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

Kinematics
In the study of kinematics, we analyze the motion of objects without focusing on the forces causing such motion. When assessing how sand falls, kinematics helps us understand its velocity just before impact. Here, we utilize the kinematic equation: \[ v = \sqrt{2gh} \]This formula enables us to calculate the velocity (\(v\)) of the sand just as it hits the dump truck. The variables in this equation include
  • \(g = 9.8 \text{ m/s}^2\), which is the constant acceleration due to gravity, and
  • \(h = 2.0 \text{ m}\), representing the height from which the sand falls.
By substituting these values into the equation, you can solve for the velocity, which is pivotal as it contributes to the subsequent momentum analysis. Understanding these basics of kinematics is crucial for solving related physics problems. Every aspect of motion, from free fall to projectile path, begins with these fundamental calculations.
Momentum
Momentum is a core concept in physics, representing the quantity of motion an object has. It is defined as the product of an object's mass and velocity. In mathematical terms, momentum \( p \) is given by:\[ p = mv \]where \( m \) is the mass and \( v \) is the velocity of the object.In this particular problem, momentum is related to how the consistently falling sand hits and accumulates momentum on the truck bed. Momentum is not only about achieving speed but also about direction and mass. In this instance, since the sand falls straight down, its direction is constant, simplifying calculations.Understanding momentum is essential for predicting how moving objects impact each other, and for explaining the rate of momentum change that directly translates to force.
Force Calculation
Force is a vector quantity that shows how much an object is pushed or pulled. In physics, it is often calculated as the rate of change of momentum. In our scenario where sand is falling vertically onto a dump truck, the force exerted by the sand can be determined using:\[ F = \frac{d}{dt}(mv) = mv \]This relates the mass flow rate of the falling sand (55.0 kg/s) and its calculated velocity. The equation implies that the force involved is directly proportional to both the mass of the sand per second and its velocity upon impact.These calculations are fundamental in physics problems involving dynamic situations where multiple forces act upon an object, influencing how it moves and interacts with other entities. Grasping force calculation helps in understanding physical interactions at both macroscopic and microscopic levels.
Impulse and Momentum
In physics, impulse is closely tied to momentum, addressing how forces affect motion over time. Impulse is defined as the change in momentum resulting from a force applied over a specific time period. Mathematically, impulse \( J \) is described by:\[ J = \Delta p = F \cdot \Delta t \]However, in this case since we are calculating per second, impulse effectively becomes the force exerted (from the falling sand), since \( \Delta t \) is 1 second.This concept is crucial for understanding how forces accumulate over time to effect substantial changes. The relationship between impulse and momentum highlights how even a brief force can lead to significant velocity changes, especially in systems where mass is constantly impacting, like the sand in this exercise.Understanding this relationship is essential for interpreting the interactions within any system where forces change over time. It shows how momentum conservation laws apply in dynamic situations, giving insight into real-world phenomena where objects collide, fall, or accelerate.

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Most popular questions from this chapter

ssm Two people are standing on a \(2.0-\mathrm{m}-\) long stationary platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a hovercraft. One person throws a \(6.0-\mathrm{kg}\) ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is 118 \(\mathrm{kg}\) . Because of the throw, this 118 -kg mass recoils. How far does it move before coming to rest again?

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A space probe is traveling in outer space with a momentum that has a magnitude of 7.5 107 kg m/s. A retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of 2.0 106 N and a direction opposite to the probe’s motion. It fires for a period of 12 s. Determine the momentum of the probe after the retrorocket ceases to fire.

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