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Chapter 3: Problem 1

The 50 -kg crate is projected along the floor with an initial speed of \(8 \mathrm{m} / \mathrm{s}\) at \(x=0 .\) The coefficient of kinetic friction is \(0.40 .\) Calculate the time required for the crate to come to rest and the corresponding distance \(x\) traveled.

Short Answer

Expert verified
The crate takes approximately 2.04 seconds to stop and travels about 8.16 meters.

Step by step solution

01

Determine the Frictional Force

The kinetic frictional force can be calculated using the equation \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force. For the crate, the normal force \( N \) is equal to its weight, which is \( m \cdot g \), where \( m = 50 \) kg and \( g = 9.81 \) m/s虏. Therefore, \( f_k = 0.40 \times 50 \times 9.81 = 196.2 \text{ N} \).
02

Apply Newton's Second Law to Determine Acceleration

Use Newton's second law \( F = m\cdot a \) to find the acceleration. The net force acting on the crate is the frictional force, which opposes the motion. Thus, \( - f_k = 50 \cdot a \). Solving for \( a \), the acceleration \( a = -\frac{196.2}{50} = -3.924 \text{ m/s}^2 \).
03

Use Kinematic Equation to Find Time to Stop

Use the kinematic equation \( v = u + at \) to find the time when velocity \( v = 0 \). Here, \( u = 8 \) m/s (initial velocity), \( a = -3.924 \) m/s虏, and \( t \) is the time. Set \( 0 = 8 + (-3.924) \cdot t \) and solve for \( t \). Thus, \( t = \frac{8}{3.924} \approx 2.04 \text{ seconds} \).
04

Use Kinematic Equation to Find Distance Traveled

Use the kinematic equation \( v^2 = u^2 + 2as \) to find the distance \( s \) traveled. Set \( 0 = 8^2 + 2(-3.924)s \) and solve for \( s \). Thus, \( s = \frac{8^2}{2 \times 3.924} \approx 8.16 \text{ meters} \).

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Key Concepts

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

Newton's Laws of Motion
Newton's laws of motion provide the foundation for understanding how forces affect motion. In this problem, we mainly use Newton鈥檚 second law, which states that the force acting on an object equals its mass multiplied by its acceleration: \[ F = m \cdot a \]
  • The force we鈥檙e concerned with here is the kinetic frictional force, which resists the motion of the crate across the floor.
  • This frictional force acts in the opposite direction of the crate's movement, slowing it down until it eventually comes to rest.
The frictional force is calculated based on the coefficient of kinetic friction (given as 0.40) and the normal force (in this case, the weight of the crate). The formula used is:\[ f_k = \mu_k \cdot N \] Once the frictional force is known, we can set it equal to the product of mass and acceleration (from Newton's second law) to solve for acceleration. This is critical in determining how quickly the object's speed changes, or its deceleration as it slides across the floor.
Kinematic Equations
Kinematic equations allow us to predict the future motion of moving objects under the influence of constant acceleration. In this scenario, we utilize these equations to calculate the time taken for the crate to stop and the distance it covers before halting. Here are the main equations used:- **To find the time to stop:** The equation used is \[ v = u + at \] where:
  • \(v\) is the final velocity (0 m/s, as the crate stops).
  • \(u\) is the initial velocity (8 m/s).
  • \(a\) is the acceleration calculated from the net force.
By rearranging the equation, we solve for time \( t \). - **To determine the distance covered:** The relevant equation is \[ v^2 = u^2 + 2as \] where \(s\) is the distance traveled. Solving for \(s\) gives us the distance from start to stop. These equations assume constant acceleration, which is valid for this problem since the only force acting is constant kinetic friction.
Acceleration
Acceleration is the rate at which an object's velocity changes over time. In this problem, the crate experiences negative acceleration, commonly referred to as deceleration, because it is slowing down.The value of this acceleration is determined by the frictional force. Here, it was calculated as \[ a = -3.924 \text{ m/s}^2 \]using a rearranged version of Newton's second law.
  • Negative acceleration indicates that the object's speed decreases over time.
  • In this scenario, the deceleration is caused by kinetic friction from the floor opposing the motion of the crate.
Understanding acceleration allows you to predict how long the object will take to stop, and how far it will travel during that time. It's a crucial aspect of analyzing motion in physics, especially when constant opposing forces like friction are involved.

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Most popular questions from this chapter

Beginning from rest when \(\theta=20^{\circ},\) a 35 -kg child slides with negligible friction down the sliding board which is in the shape of a 2.5 -m circular arc. Determine the tangential acceleration and speed of the child, and the normal force exerted on her \((a)\) when \(\theta=30^{\circ}\) and \((b)\) when \(\theta=90^{\circ}\).

Freight car \(A\) of mass \(m_{A}\) is rolling to the right when it collides with freight car \(B\) of mass \(m_{B}\) initially at rest. If the two cars are coupled together at impact, show that the fractional loss of energy equals \(m_{B} /\left(m_{A}+m_{B}\right)\).

The space shuttle launches an 800 -kg satellite by ejecting it from the cargo bay as shown. The ejection mechanism is activated and is in contact with the satellite for 4 s to give it a velocity of \(0.3 \mathrm{m} / \mathrm{s}\) in the \(z\) -direction relative to the shuttle. The mass of the shuttle is 90 Mg. Determine the component of velocity \(v_{f}\) of the shuttle in the minus \(z\) -direction resulting from the ejection. Also find the time average \(F_{\mathrm{av}}\) of the ejection force.

It is experimentally determined that the drive wheels of a car must exert a tractive force of \(560 \mathrm{N}\) on the road surface in order to maintain a steady vehicle speed of \(90 \mathrm{km} / \mathrm{h}\) on a horizontal road. If it is known that the overall drivetrain efficiency is \(e_{m}=0.70,\) determine the required motor power output \(P\).

The car is moving with a speed \(v_{0}=65 \mathrm{mi} / \mathrm{hr}\) up the 6 -percent grade, and the driver applies the brakes at point \(A,\) causing all wheels to skid. The coefficient of kinetic friction for the rain-slicked road is \(\mu_{k}=0.60 .\) Determine the stopping distance \(s_{A B} .\) Repeat your calculations for the case when the car is moving downhill from \(B\) to \(A\).
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