Chapter 9: Q. 24 (page 228)

A $150$ g particle at $x = 0$ is moving at $2.00$ m/s in the $+ x -$ direction. As it moves, it experiences a force given by $F_{x} = (0.250 N) \sin (x / 2.00 m)$ . What is the particle’s speed when it reaches $x = 3.14$ m?

Short Answer

Expert verified

Velocity of the particle at given $x$ is $3.26$ m/s

Step by step solution

Step 1. Given Information

Mass of paricle $m = 150 g = 0.15 k g$

Initial Velocity $v_{i} = 2$ m/s

Initial position $x_{i} = 0$ m

Final position $x_{f} = 3.14$ m

Variable force acting $F_{x} = (0.250 N) \sin (x / 2 m)$

Final velocity $v_{f} = ?$

Step 2. Find the velocity of a particle

We have given variable force acting on a particle and we need to find the velocity at given $x$ . We can do so by calculating work done by the force for moving particles from $x = 0$ to $x = 3.14$ m and then by equating total work done as a change in kinetic energy

$W = \int_{x_{i}}^{x_{f}} F_{x} d x = \int_{0}^{3.14} 0.250 \sin \frac{x}{2} d x = - 0.50 \cos \frac{x}{2} = 0.5 J$

Now,

$W = ∆ K W = \frac{1}{2} m {v_{f}}^{2} - \frac{1}{2} m {v_{i}}^{2} \frac{1}{2} m {v_{f}}^{2} = W + \frac{1}{2} m {v_{i}}^{2} = 0.5 + \frac{1}{2} \times 0.15 \times 2^{2} = 0.8 {v_{f}}^{2} = \frac{2 \times 0.8}{0.15} = 10.67 v_{f} = 3.26 m / s$

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A $150$ g particle at $x = 0$ is moving at $2.00$ m/s in the $+ x -$ direction. As it moves, it experiences a force given by $F_{x} = (0.250 N) \sin (x / 2.00 m)$ . What is the particle’s speed when it reaches $x = 3.14$ m?

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Find the velocity of a particle

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A 150 g particle at x=0 is moving at 2.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.250N)sin(x/2.00m). What is the particle’s speed when it reaches x=3.14 m?

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Find the velocity of a particle

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Turning Points in Physics

Linear Momentum

Work Energy and Power

Astrophysics

Dynamics

Oscillations

Study anywhere. Anytime. Across all devices.

A $150$ g particle at $x = 0$ is moving at $2.00$ m/s in the $+ x -$ direction. As it moves, it experiences a force given by $F_{x} = (0.250 N) \sin (x / 2.00 m)$ . What is the particle’s speed when it reaches $x = 3.14$ m?