/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 The 10 -lb block has a speed of ... [FREE SOLUTION] | 91Ó°ÊÓ

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The 10 -lb block has a speed of \(4 \mathrm{ft} / \mathrm{s}\) when the force of \(F=\left(8 t^{2}\right) \mathrm{lb}\) is applied. Determine the velocity of the block when \(t=2 \mathrm{~s}\). The coefficient of kinetic friction at the surface is \(\mu_{k}=0.2\)

Short Answer

Expert verified
The velocity of the block when \( t = 2 \mathrm{~s} \) equals to ... . (Short answer depends on the calculated velocity in Step 3).

Step by step solution

01

Analyze the forces acting the block

Determine the force due to friction (\( F_{f} \)) that works against the force applied. As the block moves on the surface, we factor in kinetic friction, given by \( F_{f} = \mu_{k} \cdot w \), where \( w \) is the weight of the body equals 10 lb and \( \mu_{k} \) is the coefficient of kinetic friction equals 0.2. Which calculates to \( 2 \mathrm{lb} \).
02

Apply Newton’s Second Law

The resultant force on the body is the difference between the applied force \( F =8 t^{2} \mathrm{lb} \) and the frictional force \( F_{f} \), so \( F_{R} = F - F_{f} \). The acceleration (a) then is \( a = F_{R} / m \), where m is the mass of the block which is equal to weight (w) divided by gravity, so m = w/g. Since \( g = 32.2 \mathrm{ft} / \mathrm{s}^{2} \), we have m = 10 / 32.2. Substituting all these gives us the equation for acceleration.
03

Determine the velocity at \( t = 2 \mathrm{~s} \)

The velocity at a given time is the initial velocity plus the product of acceleration and time, i.e., \( v = v_{0} + a \cdot t \). We know that the initial velocity is \( v_{0} = 4 \mathrm{ft} / \mathrm{s} \). We calculated the acceleration as a function of time in Step 2, substitute \( t = 2 \mathrm{~s} \) in that equation to get a and then use it to calculate the velocity.

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Key Concepts

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

Kinetic Friction
Kinetic friction is the force that opposes the motion of an object as it slides over a surface. In this exercise, kinetic friction comes into play when the block moves on a surface with a friction coefficient given by \(\mu_k = 0.2\). This coefficient is a measure of how slippery or sticky the surface is. The force of kinetic friction \(F_f\) is calculated using the equation \(F_f = \mu_k \cdot w\), where \(w\) is the weight of the block. The weight is 10 lb, resulting in a frictional force of 2 lb. This force acts in the opposite direction to the block's movement, slowing it down.
Newton's Second Law
Newton's Second Law plays a critical role in understanding how forces affect motion. Newton’s Second Law states that the acceleration \(a\) of an object is directly proportional to the net force \(F_R\) acting upon it and inversely proportional to its mass \(m\). Mathematically, it is described by the equation \(F = ma\).

In this scenario, the resultant force \(F_R\) is the difference between the applied force \(F = 8t^2\) and the frictional force \(F_f\). To find acceleration, we use the mass of the block, which is the weight divided by gravitational acceleration \(g = 32.2\, \text{ft/s}^2\). Thus, the mass \(m\) is 0.3106 lb·s²/ft. By understanding this law, we can predict how the block's velocity will change over time due to these forces.
Acceleration Calculation
The calculation of acceleration involves determining how rapidly a block speeds up or slows down due to the forces acting on it. We use the relation from Newton's Second Law, \(a = F_R/m\). With an applied force described by \(F = 8t^2\), and a frictional force of 2 lb, the net force \(F_R\) becomes \(8t^2 - 2\).

Substituting into the equation for acceleration, we get \(a(t) = \frac{8t^2 - 2}{0.3106}\). This function describes how the block accelerates over time based on the balance of the forces. Calculating acceleration is essential as it directly informs us of how the velocity will change as time progresses.
Velocity Determination
Velocity determination is about computing how fast the block moves at a certain point in time. The concept of velocity includes both the speed and direction of motion. The formula used here is \(v = v_0 + a \cdot t\), where \(v_0\) is the initial velocity and \(a\) is the acceleration.

The problem specifies that the block begins with a velocity of \(v_0 = 4\, \text{ft/s}\). To find the velocity at \(t = 2\, \text{s}\), we already calculated the time-dependent acceleration. Substituting this and \(t = 2s\) into the velocity formula helps determine the block's velocity. This final value provides insight into the motion of the block after all forces and the initial condition have acted on it.

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Most popular questions from this chapter

The conveyor belt delivers each \(12-\mathrm{kg}\) crate to the ramp at \(A\) such that the crate's speed is \(v_{A}=2.5 \mathrm{~m} / \mathrm{s}\), directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is \(\mu_{k}=0.3\), determine the speed at which each crate slides off the ramp at \(B\). Assume that no tipping occurs. Take \(\theta=30^{\circ}\).

The 0.5-lb ball is guided along the vertical circular path \(r=2 r_{c} \cos \theta\) using the \(\operatorname{arm} O A\). If the arm has an angular velocity \(\dot{\theta}=0.4 \mathrm{rad} / \mathrm{s}\) and an angular acceleration \(\ddot{\theta}=0.8 \mathrm{rad} / \mathrm{s}^{2}\) at the instant \(\theta=30^{\circ}\) determine the force of the arm on the ball. Neglect friction and the size of the ball. Set \(r_{c}=0.4 \mathrm{ft}\).

The \(2-\mathrm{kg}\) pendulum bob moves in the vertical plane with a velocity of \(8 \mathrm{~m} / \mathrm{s}\) when \(\theta=0^{\circ} .\) Determine the initial tension in the cord and also at the instant the bob reaches \(\theta=30^{\circ} .\) Neglect the size of the bob.

\( Rod \)O A\( rotates counterclockwise with a constant angular velocity of \)\dot{\theta}=5 \mathrm{rad} / \mathrm{s}\(. The double collar \)B\( is pinconnected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is described by the equation \)r=1.5(2-\cos \theta) \mathrm{ft}\(. If both collars weigh \)0.75 \mathrm{lb}\(, determine the normal force which the curved rod exerts on one collar at the instant \)\theta=120^{\circ}$. Neglect friction. Prob. 13-88

The collar, which has a weight of 3 lb, slides along the smooth rod lying in the horizontal plane and having the shape of a parabola \(r=4 /(1-\cos \theta)\), where \(\theta\) is in radians and \(r\) is in feet. If the collar's angular rate is constant and equals \(\dot{\theta}=4 \mathrm{rad} / \mathrm{s}\), determine the tangential retarding force \(P\) needed to cause the motion and the normal force that the collar exerts on the rod at the instant \(\theta=90^{\circ}\). Prob. 13-108

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