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Chapter 3: Problem 15

A constant force acts on a \(5.0\) kg object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0\) s. Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

Short Answer

Expert verified
The force is \(-6.7 \mathrm{~N}\), acting opposite to the direction of motion.

Step by step solution

01

Determine the change in velocity

The initial velocity of the object is given as \( v_i = 7.0 \mathrm{~m/s} \) and the final velocity \( v_f = 3.0 \mathrm{~m/s} \). To find the change in velocity, we subtract the final velocity from the initial velocity: \[ v_f - v_i = 3.0 \mathrm{~m/s} - 7.0 \mathrm{~m/s} = -4.0 \mathrm{~m/s}. \]
02

Calculate the acceleration

Using the formula for acceleration \( a = \frac{v_f - v_i}{t} \) and the information that the time taken \( t = 3.0 \mathrm{~s} \), we can substitute the values: \[ a = \frac{-4.0 \mathrm{~m/s}}{3.0 \mathrm{~s}} = -1.33 \mathrm{~m/s^2}. \] This negative value indicates that the acceleration is in the direction opposite to the initial velocity, which means it is deceleration.
03

Apply Newton's second law to find the force

Using Newton's second law \( F = ma \), where \( m = 5.0 \mathrm{~kg} \) is the mass of the object and \( a = -1.33 \mathrm{~m/s^2} \) is the acceleration, we calculate the force: \[ F = (5.0 \mathrm{~kg})(-1.33 \mathrm{~m/s^2}) = -6.65 \mathrm{~N}. \] Rounding to one decimal place, this result is \( -6.7 \mathrm{~N} \). The negative sign signifies that the force is opposite to the direction of motion, meaning it's a retarding force.

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Key Concepts

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

Force Calculation
Calculating force using Newton's second law is a fundamental skill in physics. Newton's second law tells us that the force acting on an object is the product of its mass and its acceleration. This means if you know how fast an object's speed is changing (its acceleration) and how much the object weighs (its mass), you can find out the force. The equation is given by:\[ F = ma \]where:
  • \( F \) is the force in Newtons (N).
  • \( m \) is the mass in kilograms (kg).
  • \( a \) is the acceleration in meters per second squared (m/s虏).

In our example, the force was calculated using a mass of \(5.0\) kg and an acceleration of \(-1.33\) m/s虏. Plugging these values into the formula gives:\[ F = (5.0 \text{ kg})(-1.33 \text{ m/s}^2) = -6.7 \text{ N} \]The negative sign informs us that the calculated force acts in the opposite direction to the acceleration, which in this scenario means it is working against the motion. This is known as a retarding force.
Acceleration
Acceleration describes how quickly an object speeds up or slows down. It is a vector, meaning it has both magnitude and direction. To calculate acceleration when an object changes its speed over time, use the formula:\[ a = \frac{v_f - v_i}{t} \]where:
  • \( a \) is the acceleration.
  • \( v_f \) is the final velocity.
  • \( v_i \) is the initial velocity.
  • \( t \) is the time it takes for the velocity to change.

In this exercise, the object's speed changed from \(7.0\) m/s to \(3.0\) m/s over \(3.0\) seconds. Therefore, its acceleration is calculated as:\[ a = \frac{3.0 \text{ m/s} - 7.0 \text{ m/s}}{3.0 \text{ s}} = -1.33 \text{ m/s}^2 \]The negative sign tells us the object is decelerating, meaning it is slowing down. The acceleration happens in the opposite direction to the initial motion, emphasizing that the object is losing speed.
Motion
The concept of motion involves the change in position of an object over time. Understanding motion is crucial because it helps us understand the effects of forces, such as how they can change an object's speed or direction. When evaluating motion, we often look at key characteristics like velocity, speed, and acceleration.
Velocity is a vector quantity that includes both the speed and direction, whereas speed is a scalar, having only magnitude. For our object, the initial velocity was \(7.0\) m/s, and the final velocity was \(3.0\) m/s. This decrease indicates a slowing of motion due to the negative acceleration calculated.
When a force acts on an object, it can change the object's velocity鈥攎aking it speed up, slow down, or change direction. In this problem, as the object experienced a retarding force, it slowed down, demonstrating how forces influence motion.

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Most popular questions from this chapter

A \(400-g\) block with an initial speed of \(80 \mathrm{~cm} / \mathrm{s}\) slides along a horizontal tabletop against a friction force of \(0.70 \mathrm{~N}\). \((a)\) How far will it slide before stopping? (b) What is the coefficient of friction between the block and the tabletop? (a) Take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, \(-0.70 \mathrm{~N}\). Therefore, \(\Sigma F=m a \quad\) becomes \(\quad-0.70 \mathrm{~N}=(0.400 \mathrm{~kg})(a)\) from which \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). (Notice that \(m\) is always in kilograms.) To find the distance the block slides, make use of \(_{i x}\) \(=0.80 \mathrm{~m} / \mathrm{s}, f_{x}=0\), and \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2} .\) Then \(v_{\mathrm{K}}^{2}-u_{u}^{2}=2 a x\) yields $$ x=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 a}=\frac{(0-0.64) \mathrm{m}^{2} / \mathrm{s}^{2}}{(2)\left(-1.75 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.18 \mathrm{~m} $$ (b) Because the vertical forces on the block must cancel, the upward push of the table \(F_{N}\) must equal the weight \(m g\) of the block. Then $$ \mu_{k}=\frac{\text { Friction force }}{F_{N}}=\frac{0.70 \mathrm{~N}}{(0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.178 \text { or } 0.18 $$

A cord passing over a frictionless, massless pulley has a \(4.0-\mathrm{kg}\) object tied to one end and a \(12-\mathrm{kg}\) object tied to the other. Compute the acceleration and the tension in the cord.

A 20 -kg wagon is pulled along the level ground by a rope inclined at \(30^{\circ}\) above the horizontal. A friction force of \(30 \mathrm{~N}\) opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and \((b)\) an acceleration of \(0.40 \mathrm{~m} / \mathrm{s}^{2}\) ?

A force of \(100 \mathrm{~N}\) makes an angle of \(\theta\) with the \(x\) -axis and has a scalar \(y\) -component of \(30 \mathrm{~N}\). Find both the scalar \(x\) -component of the force and the angle \(\theta\). (Remember that the number \(100 \mathrm{~N}\) has three significant figures, whereas \(30 \mathrm{~N}\) has only two.) Begin your analysis by drawing a diagram. Here the data are sketched roughly in Fig. \(3-4\). We wish to find \(F_{X}\) and \(\theta\). Since $$ \sin \theta=\frac{30 \mathrm{~N}}{100 \mathrm{~N}}=0.30 $$ \(\theta=17.46^{\circ}\), and thus, to two significant figures, \(\theta=17^{\circ}\). Then, using the \(\cos \theta\), $$ F_{x}=(100 \mathrm{~N}) \cos 17.46^{\circ}=95 \mathrm{~N} $$

A \(12.0\) -g bullet is accelerated from rest to a speed of \(700 \mathrm{~m} / \mathrm{s}\) as it travels \(20.0 \mathrm{~cm}\) in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force? [Hint: Be careful with units.
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