/*! 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 32 The \(6-\mathrm{kg}\) block is f... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 32

The \(6-\mathrm{kg}\) block is falling downward at \(v_{1}=3 \mathrm{~m} / \mathrm{s}\) when it is \(8 \mathrm{~m}\) from the sandy surface. Determine the average impulsive force acting on the block by the sand if the motion of the block is stopped in time \(1.2 \mathrm{~s}\) once the block strikes the sand. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand.

Short Answer

Expert verified
The average impulsive force acting on the block by the sand is \(15 N\).

Step by step solution

01

Calculate the Initial Momentum

Firstly, one needs to calculate the initial momentum of the block. The momentum \(p_1\) of an object could be found using the equation \(p_1 = mv_1\). So, \(p_1 = (6 kg) * (3 m/s) = 18 kg*m/s\). As the block is falling down, the momentum is negative i.e. \(p_1 = -18 kg*m/s\). The negative sign denotes direction (downward).
02

Calculate the Final Momentum

Since the block is stopped by the sandy surface, the final velocity \(v_2\) is \(0 m/s\). And the momentum \(p_2 = m * v_2 = 0 kg*m/s\).
03

Calculate Change in Momentum (Impulse)

The impulse is defined as the change in momentum. It can be calculated as \(J = p_2 - p_1 = 0 - (-18 kg*m/s) = 18 kg*m/s\).
04

Find the Average Impulsive Force

The impulsive force \(F\) is the ratio of impulse to time. Since the block stops after \(1.2 s\), this force can be calculated as \(F = J / t = 18 kg*m/s / 1.2 s = 15 N\). The positive sign indicates the force exerted by the sand is in the upward direction.

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Key Concepts

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

Momentum
Momentum is a measure of the amount of motion a moving object has. It's a fundamental concept in physics defined by the product of an object's mass and its velocity. In physics terms, momentum (\( p \) ) can be calculated using the formula:
  • \( p = m \cdot v \)
where \( m \) is the mass of the object and \( v \) is its velocity.
Momentum is a vector quantity, which means it has both direction and magnitude.
In the case of our falling block, the momentum is calculated as \( -18 \text{ kg}\cdot\text{m/s} \). The negative sign signifies that the direction of motion is downward. Understanding momentum helps us analyze how forces act upon moving objects.
Impulse
Impulse is closely related to momentum because it describes the effect of a force acting over a period of time. Impulse is essentially the change in momentum resulting from a force applied to an object.
The formula to calculate impulse (\( J \) ) is:
  • \( J = \Delta p = p_{2} - p_{1} \)
where \( \Delta p \) is the change in momentum between the initial (\( p_{1} \) ) and final momentum (\( p_{2} \) ).
In the scenario of a block striking the sand, the impulse is calculated as \( 18 \text{ kg}\cdot\text{m/s} \), indicating the change in momentum as the block stops.
Impulse-Momentum Theorem
The Impulse-Momentum Theorem establishes a direct relationship between impulse and momentum. It states that the impulse applied on an object is equal to the change in its momentum.
Mathematically, it can be expressed as:
  • \( J = F \cdot t = \Delta p \)
where \( \Delta p \) is the change in momentum, \( F \) is the average force applied, and \( t \) is the time duration for which the force is applied.
In our scenario, the block has an impulse from the sand when it stops. This impulse relates to the change in momentum giving us \( J = 18 \text{ kg}\cdot\text{m/s} \). The theorem helps in calculating the average force exerted by the sand.
Average Force
Average force is the constant force that would result in the same change in motion as the actual varying forces experienced by an object over a time period. It's defined as the impulse divided by the time duration over which the force is applied:
  • \( F = \frac{J}{t} \)
where \( J \) is the impulse and \( t \) is the time.
In the problem, the block is stopped in 1.2 seconds, meaning the average upward force from the sand is \( 15 \text{ N} \). This force ensures the block's momentum goes from \( -18 \text{ kg}\cdot\text{m/s} \) to \( 0 \text{ kg}\cdot\text{m/s} \).
Understanding average force helps us in situations where force is not constant but varies over time.

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Most popular questions from this chapter

Disk \(A\) has a mass of \(250 \mathrm{~g}\) and is sliding on a smooth horizontal surface with an initial velocity \(\left(v_{A}\right)_{1}=2 \mathrm{~m} / \mathrm{s}\). It makes a direct collision with disk \(B\), which has a mass of \(175 \mathrm{~g}\) and is originally at rest. If both disks are of the same size and the collision is perfectly elastic \((e=1)\), determine the velocity of each disk just after collision. Show that the kinetic energy of the disks before and after collision is the same.

If disk \(A\) is sliding along the tangent to disk \(B\) and strikes \(B\) with a velocity \(\mathbf{v}\), determine the velocity of \(B\) after the collision and compute the loss of kinetic energy during the collision. Neglect friction. Disk \(B\) is originally at rest. The coefficient of restitution is \(e,\) and each disk has the same size and mass \(m\).

A man kicks the 150 -g ball such that it leaves the ground at an angle of \(60^{\circ}\) and strikes the ground at the same elevation a distance of \(12 \mathrm{~m}\) away. Determine the impulse of his foot on the ball at \(A\). Neglect the impulse caused by the ball's weight while it's being kicked.

The cue ball \(A\) is given an initial velocity \(\left(v_{A}\right)_{1}=5 \mathrm{~m} / \mathrm{s}\). If it makes a direct collision with ball \(B(e=0.8)\), determine the velocity of \(B\) and the angle \(\theta\) just after it rebounds from the cushion at \(C\left(e^{\prime}=0.6\right)\). Each ball has a mass of \(0.4 \mathrm{~kg}\). Neglect their size.

A ball is thrown onto a rough floor at an angle of \(\theta=45^{\circ} .\) If it rebounds at the same angle \(\phi=45^{\circ},\) determine the coefficient of kinetic friction between the floor and the ball. The coefficient of restitution is \(e=0.6 .\) Hint: Show that during impact, the average impulses in the \(x\) and \(y\) directions are related by \(I_{x}=\mu I_{y}\). Since the time of impact is the same, \(F_{x} \Delta t=\mu F_{y} \Delta t\) or \(F_{x}=\mu F_{y}\).
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