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For example if we pull a block on an incline. Then the force of gravity i.e. $mg \sin(θ)$ will be equal to force applied so that the net work done will be zero. But if $mg \sin(θ)$ is equal to force applied to the block then how will the block move? The forces will cancel out each other.

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    $\begingroup$ Your premise "the force of gravity ... will be equal to the force applied" is true ONLY if the velocity is constant. You haven't specified anything about velocities, the (long) title notwithstanding. $\endgroup$
    – Bill N
    Oct 1, 2019 at 15:00

4 Answers 4

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This point is usually glossed over in introductory physics classes, but you're right. An object that is initially at rest at one location, and is then displaced to another location, must have had some net acceleration and deceleration over the course of the motion. But it still works out that the net work done by each force is the same.

Here's how to think about it: let's say that the $+x$ direction is up the slope. Assume friction is negligible, so that the only forces we need to worry about are the applied force (in the $+x$ direction) and the force of gravity parallel to the slope (in the $-x$ direction). Imagine that the motion happens in three phases. First, the block is accelerating in the $+x$ direction from rest; second, it travels at constant velocity up the slope; and third, it decelerates back down to rest.

  1. Acceleration: During this phase, the object is accelerating in the $+x$ direction, so there must be a net force in that direction. This means that the magnitude of the applied force must be greater than the magnitude of the force of gravity. Since these forces are applied over the same distance during this phase (by definition), the magnitude of the work done by the applied force during this phase must be greater than the magnitude of the work done by gravity.

  2. Constant velocity: During this phase, the object is not accelerating, so there must be no net force on it. This means that the magnitude of the applied force must be equal to than the magnitude of the force of gravity. Since these forces are applied over the same distance during this phase (by definition), the magnitude of the work done by the applied force during this phase must be equal to the magnitude of the work done by gravity.

  3. Deceleration: During this phase, the object is accelerating in the $-x$ direction, so there must be a net force in that direction. This means that the magnitude of the applied force must be less than the magnitude of the force of gravity. Since these forces are applied over the same distance during this phase (by definition), the magnitude of the work done by the applied force during this phase must be less than the magnitude of the work done by gravity.

The key realization here is that while the applied force does a little more (positive) work than gravity does in Phase 1, gravity does a little more (negative) work in Phase 3, and they do the equal and opposite amounts of work in Phase 2. If you were to tally up the work done by each force over all three phases, you would find that the work done by the applied force is exactly opposite the work done by gravity, even if the forces don't actually balance out moment to moment.

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But if $mg \sin(θ)$ is equal to force applied to the block then how will the block move, the forces will cancel out each other.

If an object does not move, then that implies that the forces are equal, but if the forces are equal that does not imply that the object does not move.

In particular, if the forces are equal then the acceleration is zero. This means that the object continues moving at its initial velocity, which may not be zero. In a problem like this the assumption is that there is some initial velocity which allows the object to move while the forces are balanced.

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The sum of all the forces is called the net force. If the net force is zero, it does not mean that the velocity is zero, it just means that the acceleration is zero. This can be seen from Newton's 2nd law $\mathbf{F}_{\text{net}}=m\mathbf{a}$. Since $\mathbf{F}_{\text{net}}=0$ we have $m\mathbf{a}=0$ and so $\mathbf{a}=0$.

When the change in kinetic energy is zero we have $$ \Delta KE = K_\text{f} - K_\text{i} = \frac{1}{2}mv_\text{f} - \frac{1}{2}mv_\text{i} = 0 $$ which, after multiplying by 2 and dividing by the mass, gives $v_\text{f} = v_\text{i}$. The work-energy theorem says $$ F_\text{net}s = K_\text{f} - K_\text{i} $$ but since $v_\text{f} = v_\text{i}$, the right hand side is zero, and so the left hand side, $F_\text{net}s$, is zero. This is a mathematical way to understand why the total work must be zero.

Conceptually, consider it this way: The two forces are equal in magnitude but opposite in direction, and so the magnitudes of the forces will be equal, but the displacements will be negatives of each other (to understand this, remember that the general form for work is $\mathbf{F}\cdot\mathbf{s} = Fs\cos\theta$ where $\theta$ is the angle between the vectors) and so the total work will be $W_\text{tot} = Fs - Fs = 0$.

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There are two types of friction acting on the block: Static and Kinetic (sliding).

Static friction acting up the incline prevents relative motion between the block and the incline by opposing the downward force along the incline plane. The net force is zero and the block does not move. It does this until the incline angle is steep enough that the downward force of mgsinθ exceeds the maximum possible static friction force, which is given by $f_{max}=μ_{s}mg$ where $μ_{s}$ is the coefficient of static friction and $m$ is the mass of the block.

When that happens, motion starts and is resisted by the kinetic friction force. The key now is that the kinetic friction force is generally less than the static friction force. Therefore there will be a net downward force on the block causing it to accelerate.

hope this helps.

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