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A constant force acting on a body of mass \(3.0 \mathrm{~kg}\) changes its speed from \(2.0 \mathrm{~ms}^{-1}\) to \(3.5 \mathrm{~ms}^{-1}\) in \(25 \mathrm{~s}\). The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force? (A) \(0.18 \mathrm{~N}\) (B) \(0.36 \mathrm{~N}\) (C) \(0.9 \mathrm{~N}\) (D) None of these

Short Answer

Expert verified
The magnitude of the force is \(0.18 \mathrm{~N}\) and the direction is the same as that of the initial velocity. The correct answer is (A).

Step by step solution

01

Find acceleration

We need to find acceleration (a) first, which can be calculated by subtracting the initial velocity (u) from the final velocity (v) and dividing the result by the time (t) taken for the change: \[a = \frac{v - u}{t}\] Using the given values in the problem, we have: \[a = \frac{3.5 \mathrm{~ms}^{-1} - 2.0 \mathrm{~ms}^{-1}}{25 \mathrm{~s}}\]
02

Calculate acceleration

Now, we calculate the acceleration: \[a = \frac{1.5 \mathrm{~ms}^{-1}}{25 \mathrm{~s}}\] \[a = 0.06 \mathrm{~ms}^{-2}\]
03

Apply Newton's second law

Now, we can use Newton's second law of motion to find the force (F) acting on the object, using the mass (m) and the acceleration (a): \[F = m \cdot a\] Plugging in the given mass and the calculated acceleration: \[F = 3.0 \mathrm{~kg} \cdot 0.06 \mathrm{~ms}^{-2}\]
04

Calculate the force

Finally, we calculate the force: \[F = 0.18 \mathrm{~N}\]
05

Determine the direction of the force

Since the direction of motion of the body remains unchanged, the force must be acting in the same direction as the initial velocity. Therefore, the magnitude of the force is \(0.18 \mathrm{~N}\) and the direction is the same as that of the initial velocity. So the correct answer is (A).

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Key Concepts

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

Constant Force
Understanding the concept of a constant force is crucial when dealing with problems in classical mechanics. A constant force is one that remains unchanged in both magnitude and direction over time. In the given exercise, a constant force is responsible for changing the velocity of a 3.0 kg mass from 2.0 m/s to 3.5 m/s over a span of 25 seconds.

When a constant force acts on an object, it causes a uniform acceleration, meaning that the object's velocity changes at a steady rate. This is an important aspect because it simplifies the calculations and the analysis of the object's motion. In real-world scenarios, constant forces are idealizations, but they prove to be very useful for teaching fundamental principles of motion and for setting up basic physics problems.

In applications such as space travel or the analysis of frictionless surfaces, assuming a constant force provides valuable insights into the performance and behavior of objects under such forces. It's also important to note that in a vacuum, where air resistance can be neglected, objects do experience nearly constant forces.
Acceleration Calculation
Acceleration calculation is a fundamental skill when studying motion. In the context of the problem at hand, acceleration represents the rate at which the velocity of the object changes over time. Because the force acting on the body is constant, the acceleration is constant as well. The mathematical formula for calculating acceleration is:\[a = \frac{{\text{{change in velocity}}}}{t}\bra vehiclenQual\] The acceleration calculation involves taking the difference between the final velocity (\bv) and the initial velocity (\bu), and dividing this by the time (\bt) during which the change occurred. Using the provided values, the acceleration of the body was found to be 0.06 m/s\textsuperscript{2}.

A student seeking to improve their understanding of this concept should practice with different values for velocity and time, ensuring they can identify and calculate acceleration in various scenarios. Knowing how to compute acceleration accurately allows for further calculations, such as determining the force using Newton's second law, and better understanding of an object's motion under different conditions.
Linear Motion
Linear motion is the most basic motion in physics, where an object moves in a straight line. The problem discussed here is a clear example of linear motion, as the direction of the mass's movement remains unchanged despite its changing speed. This type of motion is described by several parameters, including distance, displacement, speed, velocity, and acceleration.

When analyzing linear motion, it's important to differentiate between speed—a scalar quantity, which is just the magnitude of motion—and velocity—a vector quantity, which includes both magnitude and direction. Acceleration also plays a key role; in linear motion, it can be uniform (constant) or variable (changing).

For a student trying to deepen their physics knowledge, exploring different kinds of linear motion problems, such as objects dropping in free fall, projectiles launched at an angle, or cars accelerating down a straight road, can enhance their ability to interpret and predict motion from equations and force considerations. A constant force leads to constant acceleration, which is directly proportional to the force and inversely proportional to the mass of the object, as per Newton's second law.

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Most popular questions from this chapter

A box of mass \(8 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(\theta\). Its downward motion can be prevented by applying an upward pull \(F\) and it can be made to slide upwards by applying a force \(2 F\). The co-efficient of friction between the box and the inclined plane is (A) \(\frac{1}{3} \tan \theta\) (B) \(3 \tan \theta\) (C) \(\frac{1}{2} \tan \theta\) (D) \(2 \tan \theta\)

A block of mass \(1 \mathrm{~kg}\) is placed on a rough horizontal surface connected by a light string passing over two smooth pulleys as shown. Another block of \(1 \mathrm{~kg}\) is connected to the other end of the string. The acceleration of the system is (co-efficient of friction \(\mu=0.2\) ) (A) \(0.8 \mathrm{~g}\) (B) \(0.4 \mathrm{~g}\) (C) \(0.5 \mathrm{~g}\) (D) Zero

The angular velocity of a wheel increases from 1200 rpm to 4500 rpm in \(10 \mathrm{~s}\). The number of revolutions made during this time is (A) 950 (B) 475 (C) \(237.5\) (D) \(118.75\)

A ball is travelling with uniform translatory motion. This means that (A) it is at rest. (B) the path can be a straight line or circular and the ball travels with uniform speed. (C) all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant. (D) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

A ball of mass \(1 \mathrm{~kg}\) is at rest in position \(P\) by means of two light strings, \(O P\) and \(R P\). The string \(R P\) is now cut and the ball swings to position \(Q\). If \(\theta=45^{\circ}\). Find the ratio of tensions in the strings in positions \(O P\) (when \(R P\) was not cut) and \(O Q\) (when \(R P\) was cut). Taking \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) (A) 1 (B) 2 (C) 3 (D) \(1.5\)

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