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: 200Input : 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 donefloat workDone(float F, float D){ // Stores the work done float W; W = F * D; return W;}// Function to calculate power consumedfloat power(float F, float D, float T){ // Stores the amount // of power consumed float P; P = (F * D) / T; return P;}// Driver Codeint 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 donestatic float workDone(float F, float D){ // Stores the work done float W; W = F * D; return W;}// Function to calculate power consumedstatic float power(float F, float D, float T){ // Stores the amount // of power consumed float P; P = (F * D) / T; return P;}// Driver codepublic 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 donedef workDone(F, D): return F * D# Function to calculate power consumeddef power(F, D, T): return ((F * D) / T)# Driver codeF = 100D = 20T = 100print(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 donestatic float workDone(float F, float D){ // Stores the work done float W; W = F * D; return W;}// Function to calculate power consumedstatic float power(float F, float D, float T){ // Stores the amount // of power consumed float P; P = (F * D) / T; return P;}// Driver codestatic 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 donefunction workDone(F, D){ // Stores the work done var W; W = F * D; return W;}// Function to calculate power consumedfunction power(F, D, T){ // Stores the amount // of power consumed var P; P = (F * D) / T; return P;}// Driver codevar 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> |
2000 20
Time Complexity: O(1)
Auxiliary Space: O(1)
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