Calculate Work Done and Power Consumed by a particle

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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)

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