The position of a particle moving on a stringht line under rthe action of a force is given as `x=50t-5t^(2)` Here, x is in metre and t is in second. If mass of the particle is 2 kg, the work done by the force acting on the particle in first 5 s is.

Options

A. 2500 J
B. `-2500J`
C. 5000 J
D. `-5000J`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the position function
The position of the particle is given by the equation:
\[ x = 50t - 5t^2 \]
where \( x \) is in meters and \( t \) is in seconds.

### Step 2: Calculate the velocity
The velocity \( v \) of the particle is the derivative of the position \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(50t - 5t^2) \]
Calculating this derivative:
\[ v = 50 - 10t \]

### Step 3: Calculate the acceleration
The acceleration \( a \) of the particle is the derivative of the velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(50 - 10t) \]
Calculating this derivative:
\[ a = -10 \, \text{m/s}^2 \]

### Step 4: Calculate the force acting on the particle
Using Newton's second law, the force \( F \) acting on the particle can be calculated as:
\[ F = ma \]
Given that the mass \( m = 2 \, \text{kg} \) and \( a = -10 \, \text{m/s}^2 \):
\[ F = 2 \times (-10) = -20 \, \text{N} \]

### Step 5: Calculate the work done by the force
The work done \( W \) by the force over a time interval can be calculated using the formula:
\[ W = \int F \, dx \]
We can express this in terms of time as:
\[ W = \int F \cdot v \, dt \]
Substituting the values of \( F \) and \( v \):
\[ W = \int_{0}^{5} (-20)(50 - 10t) \, dt \]

### Step 6: Evaluate the integral
Now we will evaluate the integral:
\[ W = -20 \int_{0}^{5} (50 - 10t) \, dt \]
Calculating the integral:
\[ W = -20 \left[ 50t - 5t^2 \right]_{0}^{5} \]
Evaluating at the limits:
\[ W = -20 \left[ (50 \times 5 - 5 \times 5^2) - (50 \times 0 - 5 \times 0^2) \right] \]
\[ W = -20 \left[ 250 - 125 \right] \]
\[ W = -20 \times 125 = -2500 \, \text{J} \]

### Final Answer
The work done by the force acting on the particle in the first 5 seconds is:
\[ W = -2500 \, \text{J} \]

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