/*! 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 27 The \(20-\mathrm{kg}\) crate is ... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 27

The \(20-\mathrm{kg}\) crate is lifted by a force of \(F=\left(100+5 t^{2}\right) \mathrm{N},\) where \(t\) is in seconds. Determine the speed of the crate when \(t=3 \mathrm{~s}\), starting from rest.

Short Answer

Expert verified
The speed of the crate when \(t=3 \mathrm{s}\), starting from rest is \(17.25 \mathrm{m/s}\)

Step by step solution

01

Determine the acceleration

Apply Newton鈥檚 second law, which states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. As \(F=ma\), acceleration \(a=F/m\) where \(a\) is acceleration, \(F\) is force and \(m\) is mass. According to the problem, \(F=(100+5t^2) N\) and \(m= 20kg\). So, acceleration \(a=(100+5t^2)/20 = 5+0.25t^2 m/s^2\)
02

Calculate the velocity

Here, as the acceleration is changing with time, we will use integral calculus to reach the required velocity. Apply the principle that velocity is the integral of acceleration with respect to time, in other words \(V = \int a dt\) where \(V\) is velocity, \(a\) is acceleration and \(t\) is time. So, if you integrate \(5+0.25t^2 dt\) from \(t=0\) to \(t=3s\), the integral of first term (5) from \(0\) to \(3\) becomes \(5t|_0^3 = 15m/s\) and the integral of the second term \(\int_0^3 0.25t^2 dt = 0.25*(1/3)*t^3|_0^3 = 0.25*(9) = 2.25m/s\)
03

Determine the total velocity

Once both the integrals from Step 2 have been solved, their results should be added together to yield the total velocity at \(t=3s\). Therefore, the total velocity is \(15m/s+2.25m/s=17.25m/s\)

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Key Concepts

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

Acceleration Calculation
Understanding how to calculate acceleration is crucial when applying Newton's Second Law. This law tells us that force (\( F \)) applied to an object results in acceleration (\( a \)), and it's dependent on the object's mass (\( m \)). Simply put, greater forces lead to greater accelerations for a given mass. The mathematical form is:
  • \( F = ma \)
  • Thus, \( a = \frac{F}{m} \)
In our given problem, the force is expressed as a function of time: \( F = 100 + 5t^2 \), where \( t \) represents time in seconds. The mass of the crate is 20 kg. By substituting these into the acceleration formula, we get:
  • \( a(t) = \frac{100 + 5t^2}{20} \)
  • \( a(t) = 5 + 0.25t^2 \) \( \text{m/s}^2 \)
This formula allows us to know exactly how the acceleration of the crate changes over time as the force varies.
Integral Calculus in Physics
When acceleration varies with time, as in our problem, determining velocity requires using integral calculus. This method lets us calculate the velocity as it accumulates over a period of time. In calculus terms, velocity (\( V \)) is the integral of acceleration (\( a \)) with respect to time (\( t \)):
  • \( V = \int a(t) \, dt \)
In this instance, we compute the integral of \( 5 + 0.25t^2 \) from \( t = 0 \) to \( t = 3 \) seconds:
  • \( \int_0^3 5 \, dt = 5t \Big|_0^3 = 15 \) \( \text{m/s} \)
  • \( \int_0^3 0.25t^2 \, dt = 0.25 \times \frac{1}{3} t^3 \Big|_0^3 = 2.25 \) \( \text{m/s} \)
By calculating these integrals, we capture how the velocity of the crate gains in magnitude as time progresses, considering the acceleration it experiences.
Velocity Determination
The process of determining velocity often involves combining the results from the integral of acceleration. In our case, once we've handled the complex task of integrating the time-dependent acceleration function, we simply sum the results to find the required velocity at the desired time point.
For the crate:
  • The velocity contribution from the constant acceleration term was 15 m/s.
  • The varying acceleration term added an extra 2.25 m/s.
Adding these components together gives us the total velocity at \( t = 3 \) seconds:
  • Total Velocity \( = 15 \text{ m/s} + 2.25 \text{ m/s} = 17.25 \text{ m/s} \)
This solution demonstrates how physics builds velocity from varying forces acting over time.

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Most popular questions from this chapter

The gauge pressure of water at \(A\) is \(150.5 \mathrm{kPa}\). Water flows through the pipe at \(A\) with a velocity of \(18 \mathrm{~m} / \mathrm{s}\) and out the pipe at \(B\) and \(C\) with the same velocity \(v\). Determine the horizontal and vertical components of force exerted on the elbow necessary to hold the pipe assembly in equilibrium. Neglect the weight of water within the pipe and the weight of the pipe. The pipe has a diameter of \(50 \mathrm{~mm}\) at \(A,\) and at \(B\) and \(C\) the diameter is \(30 \mathrm{~mm} \cdot \rho_{w}=1000 \mathrm{~kg} / \mathrm{m}^{3}\)

A hockey puck is traveling to the left with a velocity of \(v_{1}=10 \mathrm{~m} / \mathrm{s}\) when it is struck by a hockey stick and given a velocity of \(v_{2}=20 \mathrm{~m} / \mathrm{s}\) as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of \(0.2 \mathrm{~kg}\).

The cart has a mass of \(3 \mathrm{~kg}\) and rolls freely from \(A\) down the slope. When it reaches the bottom, a spring loaded gun fires a \(0.5-\mathrm{kg}\) ball out the back with a horizontal velocity of \(v_{b / c}=0.6 \mathrm{~m} / \mathrm{s}\), measured relative to the cart. Determine the final velocity of the cart.

Disk \(A\) has a mass of \(250 \mathrm{~g}\) and is sliding on a smooth horizontal surface with an initial velocity \(\left(v_{A}\right)_{1}=2 \mathrm{~m} / \mathrm{s}\). It makes a direct collision with disk \(B\), which has a mass of \(175 \mathrm{~g}\) and is originally at rest. If both disks are of the same size and the collision is perfectly elastic \((e=1)\), determine the velocity of each disk just after collision. Show that the kinetic energy of the disks before and after collision is the same.

A man kicks the 150 -g ball such that it leaves the ground at an angle of \(60^{\circ}\) and strikes the ground at the same elevation a distance of \(12 \mathrm{~m}\) away. Determine the impulse of his foot on the ball at \(A\). Neglect the impulse caused by the ball's weight while it's being kicked.
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