Analyzing the Impact of a Force on a Block: A Comprehensive Guide

Understanding how a force impacts an object is fundamental to physics. This guide will explore the situation where a 10 kg block is subjected to a 50 N horizontal force over a displacement of 200 meters. We will delve into three key aspects: the final kinetic energy, the final speed of the block, and the acceleration of the block in the horizontal direction. By applying the Work-Energy Principle and Newton's Second Law, we will derive the necessary equations and calculations to fully comprehend this physical scenario.

Understanding the Scenario

A 10 kg block is at rest on a frictionless surface. A 50 Newton (N) force is applied horizontally to this block, causing it to displace 200 meters. Our task is to determine the final kinetic energy of the block, its final speed, and its acceleration.

Final Kinetic Energy

Let's start with the Work-Energy Principle, which states that the work done on an object is equal to the change in its kinetic energy:

Calculating the Work Done

The work done on the block can be calculated by the formula:

W Fd

Substituting the given values:

W 50 N × 200 m 10000 Joules (J)

Initially, the block is at rest, so its initial kinetic energy KEi is 0 J.

The final kinetic energy KEf is then:

KEf KEi W 0 J 10000 J 10000 J

Final Speed of the Block

To find the final speed of the block, we use the kinetic energy formula:

KE (frac{1}{2}mv^2)

Since the final kinetic energy is 10000 J and the mass is 10 kg:

10000 J (frac{1}{2}) × 10 kg × (v^2)

Finding (v^2):

(10000 5v^2 implies v^2 frac{10000}{5} 2000 implies v sqrt{2000} approx 44.72 ) m/s

Acceleration of the Block

According to Newton's Second Law, the acceleration of an object is given by:

F ma

Rearranging for a (acceleration):

a (frac{F}{m})

Substituting the given values:

a (frac{50 N}{10 kg} 5) m/s2

Summary of Results

Final kinetic energy: 10000 J Final speed of the block: approximately 44.72 m/s Acceleration of the block: 5 m/s2

Additional Insights

It's worth noting that the mass of the block is irrelevant to the work done. Essentially, the energy expended (work) is calculated simply as the force multiplied by the distance:

70 N × 200 m 14000 Joules

This work is also the object's kinetic energy since there are no frictional losses involved.

Using the principle of conservation of energy, we can derive the block's velocity at the 200 meters mark. The kinetic energy Ke of an object with mass m and velocity v is given by:

Ke (frac{1}{2}mv^2)

Substituting the given values:

14000 J (frac{1}{2}) × 10 kg × (v^2)

Finding v:

(14000 5v^2 implies v^2 frac{14000}{5} 2800 implies v sqrt{2800} approx 52.915 ) m/s (5 significant figures)

Alternatively, the velocity at the 200 meters mark can also be calculated using Newton's laws of motion, which confirm the consistency of our previous results.