计算完成的功和粒子消耗的功率

给定三个整数F，D和T分别表示作用在粒子上的力，粒子行进的位移和消耗的时间，任务是计算该粒子完成的功( W )和消耗的功率( P )。

例子：

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

方法：可以使用以下公式来解决问题，以计算完成的工作量和功耗：

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)

下面是上述方法的实现：

C++14

// C++ program to implement
// the above approach
#include 
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

 
// 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
 

输出：

2000
20

时间复杂度： O(1)
辅助空间： O(1)