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Chapter 8: Problem 45

What average resisting force must act on a \(3.0\) -kg mass to reduce its speed from \(65 \mathrm{~cm} / \mathrm{s}\) to \(15 \mathrm{~cm} / \mathrm{s}\) in \(0.20 \mathrm{~s}\) ?

Short Answer

Expert verified
The average resisting force is \(-7.5\text{ N}\).

Step by step solution

01

Convert Units

The given velocities are in centimeters per second, but for consistency with standard units, we will convert them to meters per second. As such, the initial velocity \( u = 65 \text{ cm/s} \) is equivalent to \( 0.65 \text{ m/s} \) and the final velocity \( v = 15 \text{ cm/s} \) is equivalent to \( 0.15 \text{ m/s} \).
02

Use the Formula for Acceleration

The change in velocity (\( v - u \)) divided by the time taken (\( t \)) gives the acceleration \( a \). Therefore, \( a = \frac{v - u}{t} = \frac{0.15 \text{ m/s} - 0.65 \text{ m/s}}{0.20 \text{ s}} = \frac{-0.50 \text{ m/s}}{0.20 \text{ s}} = -2.5 \text{ m/s}^2 \). This results in a negative acceleration, which is expected because the object is slowing down.
03

Apply Newton's Second Law

Newton's second law states that the force \( F \) acting on an object is the product of its mass \( m \) and its acceleration \( a \). So, \( F = m \cdot a = 3.0 \text{ kg} \cdot (-2.5 \text{ m/s}^2) = -7.5 \text{ N} \).
04

Interpret the Negative Sign

The negative sign of the force indicates that it is a resisting force, acting in the opposite direction of motion to slow down the object.

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Key Concepts

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

Acceleration
Acceleration refers to the rate of change of velocity over time. It tells us how quickly an object is speeding up or slowing down. In the exercise, we calculated acceleration using the formula:
  • \[ a = \frac{v - u}{t} \]
Here, \( v \) is the final velocity, \( u \) is the initial velocity, and \( t \) is the time taken for the change. The acceleration was found to be \(-2.5 \, \text{m/s}^2\), which is negative, indicating the object is decelerating or slowing down.
This negative sign is crucial as it shows the direction of acceleration is opposite to that of motion, a key point when analyzing forces acting on objects.
Resisting Force
A resisting force opposes the motion of an object. In our scenario, the force is necessary to reduce the object's speed. Newton's Second Law provides the framework to determine this force. The law is given by:
  • \[ F = m \cdot a \]
Where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. Calculating gives us a force of \(-7.5 \, \text{N}\).
The negative sign here again indicates the nature of the force as resisting. It acts opposite to the direction of the motion to slow the object down. This concept is vital for understanding how and why objects stop or slow down in real-world scenarios.
Unit Conversion
Unit conversion plays a critical role in making calculations consistent and understandable, especially when working with physical equations. Velocity in our problem was initially given in centimeters per second but needed conversion to meters per second for compatibility with standard units used in physics. The conversion is straightforward:
  • \( 1 \, \text{cm} = 0.01 \, \text{m} \)
Applying this to the velocities:
  • Initial velocity: \( 65 \, \text{cm/s} = 0.65 \, \text{m/s} \)
  • Final velocity: \( 15 \, \text{cm/s} = 0.15 \, \text{m/s} \)
Ensuring units are consistent is crucial because it helps prevent errors in calculations and ensures meaningful results. Keeping track of units can often be the key to solving physics problems correctly.
Motion
Motion describes the change in position of an object over time and can be influenced by forces acting on the object. It encapsulates several factors, including speed, velocity, and direction.
In this exercise, the motion is influenced by a force that opposes the movement, causing the object to slow down.
Understanding motion requires considering both the current state of movement (speed and direction) and the external factors acting upon it (such as resisting forces). These concepts together help explain how objects start, stop, speed up, or slow down.

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Most popular questions from this chapter

A billiard ball moving at a speed \(u_{1 i}\) strikes, head-on, another billiard ball that is at rest. Assuming the collision is completely elastic, show that $$ v_{1 i}^{2}=v_{\mathrm{lf}}^{2}+v_{2 f}^{2} $$

A ball of mass \(m\) at rest at the coordinate origin explodes into three equal pieces. At some instant, one piece is on the \(x\) -axis at \(x=40\) \(\mathrm{cm}\) and another is at \(x=20 \mathrm{~cm}, y=-60 \mathrm{~cm}\). Where is the third piece at that instant?

Sand drops at a rate of \(2000 \mathrm{~kg} / \mathrm{min}\) from the bottom of a stationary hopper onto a belt conveyer moving horizontally at \(250 \mathrm{~m} / \mathrm{min}\). Determine the force needed to drive the conveyer, neglecting friction. [Hint: How much momentum must be imparted to the sand each second?]

What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in Fig. 8-6? The horizontal speed of the water is \(80 \mathrm{~cm} / \mathrm{s}\), and \(30 \mathrm{~mL}\) of the water hit the plate each second. Assume the water moves parallel to the plate after striking it. One milliliter (mL) of water has a mass of \(1.00 \mathrm{~g}\). This question deals with speed, mass, time, and force, and that suggests impulse-momentum and Newton's Second Law. The plate exerts an impulse on the water and changes its horizontal momentum. The water exerts a counterforce on the plate. Taking the direction to the right as positive, (Impulse) \(_{x}=\) Change in \(x\) -directed momentum $$ F_{x} \Delta t=\left(m v_{x}\right)_{\text {final }}-\left(m v_{x}\right)_{\text {initial }} $$ Let \(t\) be \(1.00 \mathrm{~s}\) so that \(m\) will be the mass that strikes in \(1.00 \mathrm{~s}\), namely \(30 \mathrm{~g}\). Then the above equation becomes $$ F_{x}(1.00 \mathrm{~s})=(0.030 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})-(0.030 \mathrm{~kg})(0.80 \mathrm{~m} / \mathrm{s}) $$ from which \(F_{x}=-0.024 \mathrm{~N}\). This is the force exerted by the plate on the water. The law of action and reaction tells us that the jet exerts an equal but opposite force on the plate.

A ball is dropped from a height \(h\) above a tile floor and rebounds to a height of \(0.65 h\). Find the coefficient of restitution between ball and floor. Assign floor quantities the subscript 1 , and ball quantities the subscript \(2 .\) The initial and final velocities of the floor, \(u_{1}\) and \(v_{1}\), are zero. Therefore, $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=-\frac{v_{2}}{u_{2}} $$ Since we know both the drop and rebound heights ( \(h\) and \(0.65 h\) ), we can write equations for the interchange of \(\mathrm{PE}_{\mathrm{G}}\) and \(\mathrm{KE}\) before and after the impact \(\sqrt{665}=0.8\) Notice that the coefficient of restitution equals the square root of the final rebound height over the initial drop height.
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