Animated Solution for Physics - Work, Energy, and Power: A block of mass $m$ is kept on a platform which starts from rest with constant acceleration $\frac{g}{2}$ upwards as shown in figure. Work done by normal reaction on block in time $t$ is

A Block on the Move

Imagine you are standing in an elevator that suddenly starts accelerating upwards. You feel heavier, right? That's exactly what the block in this problem is experiencing. We have a block of mass $m$ resting on a platform. This platform isn't just sitting there; it's accelerating upwards with a constant acceleration of $a=\frac{g}{2}$.

Our mission is to find the work done by the normal reaction on the block over a certain time $t$. To do this, we need to break the problem down into two fundamental pieces: finding the exact magnitude of the normal force, and determining how far the block has traveled in that time.

Let's start by drawing a Free Body Diagram (FBD) for the block. There are two primary forces acting on it in the vertical direction: 1. The gravitational force pulling it down, which is $mg$. 2. The normal reaction force pushing it up, which we'll call $N$.

Since the block is accelerating upwards along with the platform, the net force must be directed upwards. According to Newton's Second Law, the net force equals mass times acceleration:

We can rearrange this to solve for the normal force:

This equation beautifully explains why you feel heavier in an upward-accelerating elevator! Now, we substitute the given acceleration $a=\frac{g}{2}$ into our equation:

So, the normal force is $1.5$ times the actual weight of the block.

Next, we need to figure out how far the block has moved. The problem states that the platform starts from rest, which means the initial velocity $u=0$. We can use the second equation of kinematics to find the displacement $s$:

Plugging in our values ($u=0$ and $a=\frac{g}{2}$):

Now we have everything we need. The work done by a constant force is the dot product of the force vector and the displacement vector. Since both the normal force $N$ and the displacement $s$ are pointing in the exact same upward direction, the angle between them is $0^{\circ }$, and $\cos (0^{\circ })=1$.

Therefore, the work done is simply the product of their magnitudes:

Substitute the values we just calculated:

And there we have it! The work done by the normal reaction is $\frac{3m{g}^2{t}^2}{8}$. This is a classic example of how combining Newton's laws with basic kinematics can elegantly solve work-energy problems.