Question

A \(106-\mathrm{kg}\) object is initially moving in a straight line with a speed of \(51.3 \mathrm{~m} / \mathrm{s}\). (a) If it is brought to a stop with a deceleration of \(1.97 \mathrm{~m} / \mathrm{s}^{2}\), what force is required, what distance does the object travel, and how much work is done by the force? \((b)\) Answer the same questions if the object's deceleration is \(4.82 \mathrm{~m} / \mathrm{s}^{2}\).

Short answer

The force required to stop the object is -208 kg m/s^2 for (a) and -511 kg m/s^2 for (b). The object travels approximately 1330 m for (a) and 546 m for (b). The work done by the force is approximately -278000 Joules for (a) and -279000 Joules for (b).

Step-by-step solution

Calculate Force

The force required can be calculated by using Newton's second law of motion which states that Force \(F\)equals mass \(m\) times acceleration \(a\), \(F=ma\). In this case, the acceleration is the deceleration caused by the stopping force, which should be -1.97 m/s^2 for (a) and -4.82 m/s^2 for (b). The force required to stop the object, is thus \(F = m \cdot a = 106 \, kg \cdot -1.97 \, m/s^2\) for (a) and \(F = m \cdot a = 106 \, kg \cdot -4.82 \, m/s^2\) for (b).

Calculate Distance

The distance travelled can be calculated by using the equations of motion. By rearranging \(v^2 = u^2 - 2a s\) (where \(v\) is final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, \(s\) is the distance), we have \(s = (u^2 - v^2) / 2a\). Substituting the known values, we find the distances \(s_a = (51.3 m/s)^2 / (2 \cdot 1.97 m/s^2)\) for (a) and \(s_b = (51.3 m/s)^2 / (2 \cdot 4.82 m/s^2)\) for (b).

Calculate work done by force

The work done by a force is the product of the force and the distance over which it acts and can be given by the equation \(Work = Force \cdot Distance\). From step 1 and step 2, we can calculate the work done as \(Work_a = F_a \cdot s_a\) and \(Work_b = F_b \cdot s_b\)

Key concepts

Deceleration Calculations

Deceleration is the rate at which an object slows down. It is a form of acceleration, but instead of increasing speed, it refers to a decrease in speed.

To calculate deceleration, you can use the formula \( a = \frac{\Delta v}{\Delta t} \) where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time. A negative sign is often used to indicate that the velocity is decreasing. In our exercise, the object's initial speed is \(51.3 \,\text{m/s}\) and it comes to a stop, meaning the final speed is \(0 \,\text{m/s}\), resulting in a change of velocity of \( -51.3 \,\text{m/s}\).

Deceleration can also be directly provided, as in the example, where it is \(1.97 \,\text{m/s}^2\) for part (a) and \(4.82 \,\text{m/s}^2\) for part (b), already indicating a slowing down motion against the initial direction of travel.

Force Calculation

Force is a vector quantity that causes an object to undergo a change in speed, direction, or both. Newton's second law of motion (\( F = m \times a \) ) links force (\(F\)), mass (\(m\)), and acceleration (\(a\)).

When calculating the force that brings an object to a stop (deceleration), the acceleration here will be a negative value. For our exercise, the force needed for deceleration in part (a) is \( F = 106 \,\text{kg} \times -1.97 \,\text{m/s}^2 = -208.82 \,\text{N} \) and the force in part (b) is \( F = 106 \,\text{kg} \times -4.82 \,\text{m/s}^2 = -511.32 \,\text{N} \). The negative sign represents that the force is in the opposite direction of the initial motion.

Work Done by Force

Work is defined as the energy transferred to or from an object via the application of a force along a displacement. In its simplest form, work is given by the formula \( W = F \times d \) where \(W\) is work, \(F\) is the magnitude of the force and \(d\) is the distance moved by the force.

In the case where the object comes to a stop due to a constant deceleration, the force applied is in the direction opposite to the object's initial movement. The work done by this stopping force is thus negative, because the direction of the force is opposite to the displacement. Using the calculations from the example, the distance traveled until the object stops can be calculated and then multiplied by the stopping force to find the work.

Equations of Motion

The equations of motion are a set of formulas that predict the movement of objects. They are used when the motion involves a constant acceleration and can describe how the velocity of an object changes over time.

One of these equations is \( v^2 = u^2 + 2as \) where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is acceleration, and \(s\) is the displacement. When rearranged to solve for displacement, the formula becomes \( s = \frac{u^2 - v^2}{2a} \) which is used in our exercise to find the distance the object traveled before coming to a stop.

It's crucial to consider the sign of acceleration (deceleration in this case) when using these equations, as it affects the direction of the force and displacement.