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A body starts from rest and acquires a v...

A body starts from rest and acquires a velocit V in time T. The work done on the body is time t is proportional to

A

`V/T t`

B

`(V^(2)t^(2))/T`

C

`(V^(2)t)/(T^(2))`

D

`(V^(2)t^(2))/(T^(2))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will analyze the motion of the body and apply the relevant physics concepts. ### Step-by-Step Solution: 1. **Understanding the Problem**: - A body starts from rest, meaning its initial velocity \( u = 0 \). - It acquires a final velocity \( V \) in a time \( T \). - We need to find the work done on the body in a smaller time \( t \). 2. **Finding Acceleration**: - The acceleration \( a \) can be calculated using the formula: \[ a = \frac{\Delta v}{\Delta t} = \frac{V - 0}{T} = \frac{V}{T} \] 3. **Finding Displacement in Time \( t \)**: - The displacement \( s \) during time \( t \) can be calculated using the equation of motion: \[ s = ut + \frac{1}{2} a t^2 \] - Since \( u = 0 \), the equation simplifies to: \[ s = \frac{1}{2} a t^2 \] - Substituting \( a = \frac{V}{T} \): \[ s = \frac{1}{2} \left(\frac{V}{T}\right) t^2 = \frac{V t^2}{2T} \] 4. **Calculating the Force**: - According to Newton's second law, the force \( F \) acting on the body is: \[ F = ma \] - Substituting \( a = \frac{V}{T} \): \[ F = m \left(\frac{V}{T}\right) \] 5. **Calculating Work Done**: - The work done \( W \) is given by: \[ W = F \cdot s \] - Substituting the expressions for \( F \) and \( s \): \[ W = \left(m \frac{V}{T}\right) \left(\frac{V t^2}{2T}\right) \] - Simplifying this gives: \[ W = \frac{m V^2 t^2}{2 T^2} \] 6. **Identifying Proportionality**: - From the expression \( W = \frac{m V^2 t^2}{2 T^2} \), we can see that the work done \( W \) is proportional to \( V^2 t^2 \) when considering \( m \) and \( 2 \) as constants. - Thus, we can write: \[ W \propto \frac{V^2 t^2}{T^2} \] ### Final Answer: The work done on the body in time \( t \) is proportional to \( \frac{V^2 t^2}{T^2} \). ---
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