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Chapter 5: Problem 12

Besides its weight, a \(2.80-\mathrm{kg}\) object is subjected to one other constant force. The object starts from rest and in \(1.20 \mathrm{s}\) experiences a displacement of \((4.20 \hat{\mathbf{i}}-3.30 \hat{\mathbf{j}}) \mathrm{m}\) where the direction of \(j\) is the upward vertical direction. Determine the other force.

Short Answer

Expert verified
The other constant force acting on the object is \(\vec{F} = 8.176 \hat{\mathbf{i}} - 21.028 \hat{\mathbf{j}} N \).

Step by step solution

01

Calculate the velocity

First, calculate the velocity after 1.20 s. The velocity can be found from the displacement, which is the change in position over time. Since displacement = velocity x time, the velocity can be calculated as \(\frac{displacement}{time}\). Thus the velocity = \(\frac{(4.20 \hat{\mathbf{i}} - 3.30 \hat{\mathbf{j}}) \mathrm{m}}{1.20s}\) = \(3.5 \hat{\mathbf{i}} - 2.75 \hat{\mathbf{j}} ms^{-1}\).
02

Calculate the acceleration

Next, find the acceleration. At the start, the object is at rest so its initial velocity is 0. Acceleration = change in velocity / time. Thus acceleration = \(\frac{(3.5 \hat{\mathbf{i}} - 2.75 \hat{\mathbf{j}} ms^{-1} - 0)}{1.20s}\) = \(2.92 \hat{\mathbf{i}} - 2.29 \hat{\mathbf{j}} ms^{-2}\).
03

Apply Newton's 2nd Law

Next, apply Newton's 2nd law in vector form, \(\sum F = m*a\). The only force in the x-direction is due to the other constant force, say F. In the y-direction, the forces are the weight (mg) and also F. So in vector form, the sum of forces is: \(\sum F = F_{x} + F_{y} = m*a\), where \(F_{x} = F_{i} = m*a_{i} = 2.80 Kg*2.92 ms^{-2} = 8.176 N\) and \(F_{y} = F_{j} + weight = m*a_{j} + m*g = 2.80 Kg*(-2.29 ms^{-2}) + 2.80 Kg*9.8 ms^{-2} = 21.028 N\).
04

Determine the other force

Finally, combining the x and y components of F, we obtain the other constant Force in vector form as \(\vec{F} = 8.176 \hat{\mathbf{i}} - 21.028 \hat{\mathbf{j}} N \).

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Key Concepts

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

Vector Forces
In physics, a force is often described using vectors. Vectors are quantities that have both a magnitude and a direction. This makes forces particularly suitable for vector representation, as they act in specific directions with a certain strength.
This feature is crucial when dealing with multiple forces acting on an object.
  • In the x-direction, the force is denoted as \( F_x = F_i \), where \( F_i \) influences the object horizontally.
  • In the y-direction, the force is \( F_y \), and it includes other factors like weight, which acts downward due to gravity.
Each vector component can be thought of as pushing or pulling the object in its respective direction, making it simpler to analyze complex force interactions. By breaking down the forces into vector components, solutions like finding the net force become manageable using Newton's Second Law.
Displacement and Velocity
Displacement and velocity are integral to understanding motion. Displacement is the vector representation of how far and in what direction an object's position has changed. It differs from total distance covered, as it takes into account only the shortest path between initial and final positions.
Velocity is the rate at which an object's displacement changes over time, also a vector quantity, illustrating both speed and the direction of movement.
  • For our exercise, the displacement is \((4.20 \hat{\mathbf{i}} - 3.30 \hat{\mathbf{j}}) \mathrm{m}\).
  • The velocity after 1.2 seconds is calculated as \( \frac{displacement}{time} \), resulting in \(3.5 \hat{\mathbf{i}} - 2.75 \hat{\mathbf{j}} \ms^{-1}\).
Understanding these concepts is crucial for calculating further motion factors like acceleration. By expressing these as vectors, you can precisely map an object's journey through space.
Acceleration Calculation
Acceleration is the vector rate of change of an object's velocity. Calculating it involves determining how quickly an object speeds up or slows down, and in what direction.
For an object starting from rest, acceleration is straightforward to find as the velocity change over time.
  • In the problem, acceleration was calculated from the change in velocity over \(1.20 \mathrm{s}\). Initial velocity was \(0\), leading to an acceleration vector of \(2.92 \hat{\mathbf{i}} - 2.29 \hat{\mathbf{j}} \mathrm{ms^{-2}}\).
This acceleration plays a crucial role in applying Newton's Second Law, \(\sum F = m \cdot a\). Once acceleration is known, calculating the resulting forces becomes feasible, allowing the assessment of all acting forces within the scenario.

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Most popular questions from this chapter

You stand on the seat of a chair and then hop off. (a) During the time you are in flight down to the floor, the Earth is lurching up toward you with an acceleration of what order of magnitude? In your solution explain your logic. Model the Earth as a perfectly solid object. (b) The Earth moves up through a distance of what order of magnitude?

The largest-caliber antiaircraft gun operated by the German air force during World War II was the 12.8 -cm Flak 40\. This weapon fired a 25.8 -kg shell with a muzzle speed of \(880 \mathrm{m} / \mathrm{s} .\) What propulsive force was necessary to attain the muzzle speed within the 6.00 -m barrel? (Assume the shell moves horizontally with constant acceleration and neglect friction.)

A force \(\mathbf{F}\) applied to an object of mass \(m_{1}\) produces an acceleration of \(3.00 \mathrm{m} / \mathrm{s}^{2} .\) The same force applied to a second object of mass \(m_{2}\) produces an acceleration of \(1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the value of the ratio \(m_{1} / m_{2} ?\) (b) If \(m_{1}\) and \(m_{2}\) are combined, find their acceleration under the action of the force \(\mathbf{F}\).

A frictionless plane is \(10.0 \mathrm{m}\) long and inclined at \(35.0^{\circ} .\) A sled starts at the bottom with an initial speed of \(5.00 \mathrm{m} / \mathrm{s}\) up the incline. When it reaches the point at which it momentarily stops, a second sled is released from the top of this incline with an initial speed \(v_{i} .\) Both sleds reach the bottom of the incline at the same moment. (a) Determine the distance that the first sled traveled up the incline. (b) Determine the initial speed of the second sled.

The distance between two telephone poles is \(50.0 \mathrm{m}\) When a \(1.00-\mathrm{kg}\) bird lands on the telephone wire midway between the poles, the wire sags \(0.200 \mathrm{m}\). Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.
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