Introduction

This article delves into the physics behind a specific scenario involving a car. We will explore the work done on the car when a force of 2000 N is applied, considering the effect of friction. Our analysis will be split into two parts: the first 10 seconds when the car is accelerating, and the remaining 5 seconds where it is slowing down. We will employ both traditional and simplified methods to ensure a thorough understanding of the process.

Problem Description

A car with a mass of 1.5 tons (1500 kg) is subjected to a force of 2000 N. The frictional force is 500 N. The engine stops after 10 seconds. We aim to calculate the work done in 15 seconds.

Interval 1: Car Accelerating from 0 to 10 Seconds

1.1 Calculating Acceleration

Using Newton's second law, the net force (2000 N - 500 N) produces an acceleration of 1.0 m/s2.

1.2 Distance Traveled in 10 Seconds

Using the formula for distance in uniformly accelerated motion:

$$S frac{1}{2} a t^2$$

Substituting the values:

$$S 0 frac{1}{2} cdot 1.0 cdot 10^2 50 text{ meters}$$

1.3 Work Done Using Force and Distance

Calculating the work done:

$$Work F_{x} cdot d (2000 text{ N} - 500 text{ N}) times 50 text{ m} 75000 text{ Nm}$$

1.4 Work Done Using Kinetic Energy

After 10 seconds, the final velocity is calculated using the formula:

$$V_f^2 - V_i^2 2aS$$

Substituting the values:

$$10^2 - 0 2 cdot 1.0 cdot 50$$

Simplifying:

$$V_f 10 text{ m/s}$$

Calculating the kinetic energy (KE) at the end of 10 seconds:

$$KE frac{1}{2}mv^2 frac{1}{2} cdot 1500 text{ kg} cdot 10^2 75000 text{ Nm}$$

Interval 2: Car Slowing Down from 10 to 15 Seconds

2.1 Calculating Deceleration Due to Friction

Using Newton's second law, the frictional force (500 N) produces a deceleration of 0.333 m/s2.

2.2 Distance Traveled in the Last 5 Seconds

Using the formula for distance in uniformly decelerated motion:

$$S V_it frac{1}{2} a t^2$$

Substituting the values:

$$S 10 cdot 5 frac{1}{2} cdot -0.333 cdot 5^2 46.25 text{ meters}$$

2.3 Work Done by Friction

Calculating the work done by friction:

$$Work F cdot d 500 text{ N} cdot 46.25 text{ m} 23125 text{ Nm}$$

Net Work Done in 15 Seconds

The net work done is the total work done by the engine minus the work done by friction. This is calculated as:

Net Work 75000 Nm - 23125 Nm 51875 Nm

Alternative Approach for Simplicity

A more straightforward method involves using the following equations:

$$a frac{F}{m} frac{1500 text{ N}}{1500 text{ kg}} 1 text{ m/s}^2$$

After 10 seconds, the final velocity (Vf) is:

$$V_f V_i at 0 1 cdot 10 10 text{ m/s}$$

When the engine stops, the car decelerates at (frac{1}{3}) of the initial acceleration. After 15 seconds, the final velocity is:

$$V_f' V_f - frac{1}{3}t 10 - frac{1}{3} cdot 5 8.33 text{ m/s}$$

Calculating the kinetic energy at the end of 15 seconds:

$$KE frac{1}{2} cdot 1500 text{ kg} cdot (8.33)^2 52000 text{ Nm}$$

Conclusion

Both methods lead to the same result, demonstrating the consistency and accuracy of the analysis. Understanding the work done in such scenarios is crucial for advanced physics applications and real-world engineering problems.