A block of mass M is kept on a platform which starts accelerating upwards from rest with a constant acceleration a. During the time interval T, the work done by contact froce on mass M is
` (##NAR_NEET_PHY_XI_P2_C06_E09_011_Q01##) `

Options

A. `1/2 Ma^(2)T^(2)`
B. zero kinetic energy and finite momentum
C. `1/2M(g + a)^(2) T^(2)`
D. `1/2M(g - a)^(2) T^(2)`

Explanation

To solve the problem of finding the work done by the contact force on a block of mass M kept on a platform that accelerates upwards with a constant acceleration a, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
When the platform accelerates upwards, two forces act on the block:
1. The gravitational force acting downwards, which is \( F_g = mg \).
2. The normal force \( N \) exerted by the platform acting upwards.

### Step 2: Determine the Normal Force
Since the platform is accelerating upwards, we need to consider the pseudo force acting on the block due to the upward acceleration. The effective force balance can be written as:
\[ N - mg = ma \]
Rearranging this gives us:
\[ N = mg + ma = m(g + a) \]

### Step 3: Calculate the Displacement of the Block
The displacement \( s \) of the block during the time interval \( T \) can be calculated using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \) (the platform starts from rest), the equation simplifies to:
\[ s = \frac{1}{2} a T^2 \]

### Step 4: Calculate the Work Done by the Contact Force
The work done \( W \) by the contact force (normal force) on the block can be calculated using the formula:
\[ W = N \cdot s \]
Substituting the values of \( N \) and \( s \):
\[ W = (m(g + a)) \cdot \left(\frac{1}{2} a T^2\right) \]
Thus, the expression for the work done becomes:
\[ W = \frac{1}{2} m(g + a) a T^2 \]

### Final Answer
The work done by the contact force on mass \( M \) during the time interval \( T \) is:
\[ W = \frac{1}{2} m(g + a) a T^2 \]