A particle of mass m initially moving with speed v.A force acts on the particle f=kx where x is the distance travelled by the particle and k is constant. Find the speed of the particle when the work done by the force equals W.

Options

A. `sqrt((k)/(m)+v^(2))`
B. `sqrt((2W)/(m)+v^(2))`
C. `sqrt((2W)/(k)+v^(2))`
D. `sqrt((W)/(2m)+v^(2))`

Answer

To solve the problem, we need to find the speed of a particle of mass \( m \) when the work done by the force \( F = kx \) equals \( W \). Here are the steps to derive the solution:

### Step 1: Understand the Work Done by the Force
The work done \( W \) by a variable force can be expressed as:
\[ W = \int F \, dx \]
In this case, the force \( F \) is given by \( F = kx \). Therefore, the work done as the particle moves from \( 0 \) to \( x \) is:
\[ W = \int_0^x kx' \, dx' = \left[ \frac{k}{2} (x')^2 \right]_0^x = \frac{k}{2} x^2 \]

### Step 2: Relate Work Done to the Given Work \( W \)
Setting the expression for work done equal to \( W \):
\[ W = \frac{k}{2} x^2 \]
From this, we can solve for \( k \):
\[ k = \frac{2W}{x^2} \]

### Step 3: Use the Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} mv^2 \]
The final kinetic energy \( KE_f \) when the speed is \( v_f \) is:
\[ KE_f = \frac{1}{2} mv_f^2 \]
The work done \( W \) is equal to the change in kinetic energy:
\[ W = KE_f - KE_i \]
Substituting the expressions for kinetic energy:
\[ W = \frac{1}{2} mv_f^2 - \frac{1}{2} mv^2 \]
Rearranging this gives:
\[ \frac{1}{2} mv_f^2 = W + \frac{1}{2} mv^2 \]

### Step 4: Solve for Final Speed \( v_f \)
Multiplying through by 2 to eliminate the fraction:
\[ mv_f^2 = 2W + mv^2 \]
Now, divide by \( m \):
\[ v_f^2 = \frac{2W}{m} + v^2 \]
Taking the square root gives us the final speed:
\[ v_f = \sqrt{v^2 + \frac{2W}{m}} \]

### Final Answer
The speed of the particle when the work done by the force equals \( W \) is:
\[ v_f = \sqrt{v^2 + \frac{2W}{m}} \]