Animated Solution for Physics - Laws of Motion: An object of mass $m$ is being moved with a constant velocity under the action of an applied force of $2\text{ N}$ along a frictionless surface with following surface profile. The correct applied force versus distance graph will be

Imagine you are pushing a heavy cart over a perfectly smooth, frictionless hill. The path consists of a steep climb, a flat plateau at the top, and a steep descent on the other side. Your goal is to keep the cart moving at a perfectly constant velocity the entire time. How hard do you need to push or pull at each stage?

This problem is a beautiful application of Newton's First Law of Motion, which states that an object will maintain a constant velocity if and only if the net force acting on it is zero ($F_{\text{net}}=0$). Let's break down the journey into three distinct phases.

The Upward Climb

As you push the block up the first incline, gravity is working against you. The weight of the block ($mg$) acts straight down, but it has a component that points directly down the slope, given by $mg\sin \theta$.

To keep the block moving up at a constant velocity, you must apply a force $F$ that perfectly balances this downward pull. Therefore, your applied force must be directed up the slope, and its magnitude must be exactly equal to the gravitational component:

The problem tells us that this required force is $2\text{ N}$. Because you are pushing in the direction of motion, we consider this a positive force of $+2\text{ N}$.

The Flat Plateau

Once the block reaches the horizontal section at the top, the situation changes dramatically. Gravity still pulls straight down, but now the surface is completely flat. There is no component of gravity acting along the direction of motion ($mg\sin 0^{\circ }=0$).

Since the surface is frictionless, there is nothing trying to slow the block down. According to Newton's First Law, the block will just keep coasting on its own! You don't need to push it at all. Therefore, the applied force drops to zero:

The Downward Descent

Finally, the block begins its descent down the second incline. Now, gravity is pulling it down the slope with a force of $mg\sin \theta$. If you were to let go, the block would accelerate rapidly down the hill.

But remember, your job is to keep the velocity constant. To prevent the block from speeding up, you must pull backward on it, acting like a brake. Your applied force must point up the slope, opposite to the direction of motion, to perfectly cancel out gravity:

Since the angle $\theta$ is the same as before, the magnitude of the gravitational pull is still $2\text{ N}$. Because you are pulling opposite to the direction of motion, this is a negative force of $-2\text{ N}$.

If we plot your applied force $F$ against the distance $x$ traveled, we get a very distinct shape: 1. A positive rectangular pulse of $+2\text{ N}$ during the upward climb. 2. A flat line at $0\text{ N}$ across the plateau. 3. A negative rectangular pulse of $-2\text{ N}$ during the descent.

This sequence perfectly matches the graph shown in Option (b).

Adding Friction

What if this hill wasn't a frictionless physics wonderland? If kinetic friction (${\mu }_k$) were present, it would always oppose the direction of motion.

On the way up, you would have to fight both gravity and friction, requiring a much larger positive force ($F=mg\sin \theta +{\mu }_{k}mg\cos \theta$). On the way down, friction would actually help you brake, meaning you wouldn't have to pull back as hard ($F=-mg\sin \theta +{\mu }_{k}mg\cos \theta$). The symmetry of our perfect graph would be broken, shifting the entire curve upwards!