A \(5.5-\mathrm{kg}\) block is initially at rest on a frictionless horizontal surface. It is pulled with a constant horizontal force of \(3.8 \mathrm{~N}\). (a) What is its acceleration? (b) How long must it be pulled before its speed is \(5.2 \mathrm{~m} / \mathrm{s} ?(c)\) How far does it move in this time?

Short Answer

Expert verified
The acceleration of the block is given by \(\frac{F}{m}\), the time taken to reach the speed is \(\frac{v-u}{a}\), and the distance moved in this time is calculated using \(S = u \cdot t + \frac{1}{2} \cdot a \cdot t^2\). Fill in all the known values to get the final results.

Step by step solution

01

Calculate Acceleration

Start by using Newton's second law \(F = m \cdot a\) to find the acceleration. Here, the force \(F = 3.8N\) and mass \(m = 5.5kg\). So, the acceleration \(a\) can be calculated as \[a = \frac{F}{m} = \frac{3.8N}{5.5kg}.\]
02

Compute Time

Now, the time taken \((t)\) to reach the speed \(5.2 m/s\) can be calculated by rearranging the formula for acceleration: \(a = \frac{v-u}{t}\). The initial velocity \(u\) is \(0 m/s\) (since the block starts from rest) and the final velocity \(v\) is \(5.2 m/s\). Substituting the known values, we get: \[t = \frac{v-u}{a} = \frac{5.2m/s-0m/s}{a}\]
03

Find Distance

Lastly, the distance \(S\) travelled by the block can be calculated by using the formula: \(S = u \cdot t + \frac{1}{2} \cdot a \cdot t^2\). Again, the initial velocity \(u = 0 m/s\) (block starts from the rest), the acceleration \(a\) (calculated in Step 1), and the time \(t\) (calculated in step 2), we get: \[S = 0 m/s \cdot t + \frac{1}{2} \cdot a \cdot t^2\]

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Key Concepts

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

Physics Acceleration Calculation
Understanding acceleration in physics is crucial for solving problems involving motion. Acceleration is defined as the rate at which the velocity of an object changes with time. It's not just about an object speeding up; an object can also be decelerating, meaning its speed is decreasing, yet it is still considered to be accelerating in the context of physics.

To calculate acceleration, the formula from Newton's second law of motion, \( F = m \cdot a \), is often used, where \(F\) is the force applied on the object, \(m\) is the mass of the object, and \(a\) is the acceleration. Rearranging the equation for acceleration gives us \( a = \frac{F}{m} \). This formula shows that the acceleration of an object is directly proportional to the force applied and inversely proportional to the object’s mass.

In our exercise example, to find the acceleration of a block with a mass of \(5.5\mathrm{kg}\) being pulled by a force of \(3.8\mathrm{N}\), we divide the force by the mass, resulting in an acceleration value.
Force and Motion
The relationship between force and motion forms the foundation of Newton's laws of motion, particularly the second law. It tells us how the velocity of an object changes when it is subjected to a force. The second law states that the force exerted on an object is equal to the product of its mass and acceleration \( (F = m \cdot a) \).

This law provides insight into how objects behave when forces are applied. For example, a heavier object will require more force to achieve the same acceleration as a lighter one. It's important to note that the force and acceleration vectors have the same direction.

When we apply this law to the 5.5-kg block in the exercise, it helps us understand why the block starts to move and accelerates when the force is applied. It also helps to predict how the motion will continue under constant force and can even determine the effects if other forces, such as friction or air resistance, were present.
Kinematic Equations
Kinematic equations are a set of formulas that relate the five kinematic variables: displacement, initial velocity, final velocity, acceleration, and time. These equations are indispensable tools for solving constant-acceleration problems in physics. Even though the actual motion might be very complex, kinematic equations simplify the analysis to a manageable level by the assumption that acceleration remains constant.

One of the fundamental kinematic equations is \( S = u \cdot t + \frac{1}{2} \cdot a \cdot t^2 \), where \(S\) is the distance traveled, \(u\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration. Since our block starts at rest, the initial velocity \(u\) is zero, which simplifies our distance calculation.

By applying the kinematic equations to the block in our exercise, we can predict its speed after a certain time, as well as the distance the block covers in that time. These equations are a powerful tool that lets us extrapolate the future position and velocity of an object from its current state, assuming the acceleration is known and constant.

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Most popular questions from this chapter

An electron travels in a straight line from the cathode of a vacuum tube to its anode, which is \(1.5 \mathrm{~cm}\) away. It starts with zero speed and reaches the anode with a speed of \(5.8 \times\) \(10^{6} \mathrm{~m} / \mathrm{s}\). Assume constant acceleration and compute the force on the electron. This force is electrical in origin. The electron's mass is \(9.11 \times 10^{-31} \mathrm{~kg}\).

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An electron is projected horizontally at a speed of \(1.2 \times\) \(10^{7} \mathrm{~m} / \mathrm{s}\) into an electric field that exerts a constant vertical force of \(4.5 \times 10^{-16} \mathrm{~N}\) on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg} .\) Determine the vertical distance the electron is deflected during the time it has moved forward \(33 \mathrm{~mm}\) horizontally.

A certain particle has a weight of \(26.0 \mathrm{~N}\) at a point where the acceleration due to gravity is \(9.80 \mathrm{~m} / \mathrm{s}^{2} .(a)\) What are the weight and mass of the particle at a point where the acceleration due to gravity is \(4.60 \mathrm{~m} / \mathrm{s}^{2} ?(b)\) What are the weight and mass of the particle if it is moved to a point in space where the gravitational force is zero?

A neutron travels at a speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s} .\) Nuclear forces are of very short range, being essentially zero outside a nucleus but very strong inside. If the neutron is captured and brought to rest by a nucleus whose diameter is \(1.0 \times\) \(10^{-14} \mathrm{~m}\), what is the minimum magnitude of the force, presumed to be constant, that acts on this neutron? The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\)

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