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Chapter 13: Problem 45

If the force exerted on cable \(A B\) by the motor is \(F=\left(100 t^{3 / 2}\right) \mathrm{N}\), where \(t\) is in seconds, determine the \(50-\mathrm{kg}\) crate's velocity when \(t=5 \mathrm{~s}\). The coefficients of static and kinetic friction between the crate and the ground are \(\mu_{s}=0.4\) and \(\mu_{k}=0.3\), respectively. Initially the crate is at rest. Prob. 13-45

Short Answer

Expert verified
The crate's velocity when t=5s is approximately \( 18.63 m/s \).

Step by step solution

01

Analyze the Forces

Begin by examining the forces at play. \( F = 100t^{3/2} \) is the force exerted by the motor on the cable and the friction force can be calculated with the formula \( F_f = \mu_k * N \) where \( N = m*g \) is the weight of the crate (m is its mass and g represents the gravity constant, \( 9.81 m/s^2 \)).
02

Calculate the Frictional Force

Using the given mass m=50 kg and 碌k = 0.3, calculate the kinetic friction force. Here, \( N = m * g = 50 * 9.81 = 490.5 N \), and thus the friction force \( F_f = \mu_k * N = 0.3 * 490.5 = 147.15 N \)
03

Determine the Net Force

The net force acting on the object can be obtained from the difference between the force exerted by the motor and the frictional force using the formula \( F_{net}=F-F_f \).
04

Solve for the Velocity

Using the equation for force \( F = ma \), where m is mass and a is acceleration, solve for acceleration \( a = F_{net} / m \). Then, calculate the velocity using \( v = u + at \), where u is initial velocity, a is acceleration and t is time. As the crate was initially at rest, u = 0. Substitute time = 5 seconds into the force equation and use the calculated acceleration to find the velocity.

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Key Concepts

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

Force Exerted by the Motor
Let's begin by understanding the force exerted by the motor on the cable. This force is described by the formula \( F = 100 t^{3/2} \), where \( t \) represents the time in seconds. This equation tells us that the force is not just a constant value; it actually depends on time and increases as time increases. At any given second, you can determine the exact force by plugging the time into the equation. For example, at \( t = 5 \) seconds, calculate the force as follows:
  • Plug \( t = 5 \) into the equation, giving \( F = 100 \times (5)^{3/2} \)
  • Simplifying gives \( F = 100 \times 11.18 = 1118 \) N
The motor thus exerts 1118 N on the cable at 5 seconds. As time changes, so does this force.
Understanding Net Force
Net force is crucial in determining motion, as it represents the total force acting on an object. In this scenario, we calculate it by subtracting the opposing kinetic friction from the motor's exerted force.
Kinetic friction arises due to the contact between the moving crate and the ground and is calculated as follows:
  • The normal force \( N \) is the weight of the crate, calculated by \( N = m \times g = 50 \times 9.81 = 490.5 \) N
  • The frictional force \( F_f \) is given by \( F_f = \mu_k \times N = 0.3 \times 490.5 = 147.15 \) N
With these forces, calculate net force \( F_{net} \) with:
  • \( F_{net} = F_{exerted} - F_f = 1118 - 147.15 = 970.85 \) N
This net force will be used to find the crate's acceleration and subsequent velocity.
Velocity Calculation of the Crate
With the net force established, it's time to calculate the velocity of the crate. Use Newton's second law of motion, \( F = ma \), to find the acceleration \( a \):
  • \( a = \frac{F_{net}}{m} = \frac{970.85}{50} = 19.417 \text{ m/s}^2 \)
Next, determine the velocity with the equation \( v = u + at \), where \( u \) is the initial velocity (0 m/s since the crate is initially at rest), \( a \) is acceleration, and \( t \) is time:
  • At \( t = 5 \) seconds \( v = 0 + 19.417 \times 5 = 97.085 \text{ m/s} \)
Thus, the velocity of the crate at 5 seconds is approximately 97.085 m/s. This outcome demonstrates how the forces interact over time to alter the motion of the crate.

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Most popular questions from this chapter

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