Calculate Work Done and Power Consumed by a particle

Given three integers F, D, and T representing the force acting on a particle, displacement traveled by the particle, and time consumed respectively, the task is to calculate Work done (W) and Power consumed (P) by that particle.

Examples:

Input: F = 100, D = 20, T = 100
Output: 
Work done: 2000
Power Consumed: 200

Input : F=40.2, D=10.6, T=20
Output : 
Work Done: 426.12
Power Consumed: 21.306

Approach: The problem can be solved using the following formulas to calculate Work done and Power consumed:

Work done (W) = Force acting on particle (F) * Displacement of Particle (D)

Power Consumed (P) = Force acting on particle (F) * Displacement of Particle (D) / Time consumed (T)

Below is the implementation of the above approach:

C++14

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate work done
float workDone(float F, float D)
{
    // Stores the work done
    float W;
    W = F * D;

    return W;
}
// Function to calculate power consumed
float power(float F, float D, float T)
{
    // Stores the amount
    // of power consumed
    float P;
    P = (F * D) / T;

    return P;
}

// Driver Code
int main()
{
    float F = 100, D = 20, T = 100;

    cout << workDone(F, D) << endl;
    cout << power(F, D, T) << endl;

    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{

// Function to calculate work done
static float workDone(float F, float D)
{

    // Stores the work done
    float W;
    W = F * D;

    return W;
}

// Function to calculate power consumed
static float power(float F, float D, float T)
{

    // Stores the amount
    // of power consumed
    float P;
    P = (F * D) / T;

    return P;
}

// Driver code
public static void main(String[] args)
{
    float F = 100, D = 20, T = 100;

    System.out.println(workDone(F, D));
    System.out.println(power(F, D, T));
}
}

// This code is contributed by abhinavjain194

Python3

# Python3 program to implement
# the above approach

# Function to calculate work done
def workDone(F, D):

    return F * D

# Function to calculate power consumed
def power(F, D, T):

    return ((F * D) / T)

# Driver code
F = 100
D = 20
T = 100

print(workDone(F, D))
print(power(F, D, T))

# This code is contributed by abhinavjain194

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to calculate work done
static float workDone(float F, float D)
{

    // Stores the work done
    float W;
    W = F * D;

    return W;
}

// Function to calculate power consumed
static float power(float F, float D, float T)
{

    // Stores the amount
    // of power consumed
    float P;
    P = (F * D) / T;

    return P;
}

// Driver code
static void Main()
{
    float F = 100, D = 20, T = 100;

    Console.WriteLine(workDone(F, D));
    Console.WriteLine(power(F, D, T));
}
}

// This code is contributed by abhinavjain194

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to calculate work done
function workDone(F, D)
{

    // Stores the work done
    var W;
    W = F * D;

    return W;
}

// Function to calculate power consumed
function power(F, D, T)
{

    // Stores the amount
    // of power consumed
    var P;
    P = (F * D) / T;

    return P;
}

// Driver code
var F = 100, D = 20, T = 100;

document.write(workDone(F, D) + "<br>");
document.write(power(F, D, T));

// This code is contributed by Khushboogoyal499

</script>

Output:

2000
20

Time Complexity: O(1)
Auxiliary Space: O(1)