4. A box of mass 0.50 kg is pulled by a force of magnitude $$ F=2.0 \mathrm{~N} $$ across a rough, horizontal surface as shown. If the box moves with constant speed along the straight path, what is the total work done on the box when it covers a horizontal distance of 10 m ?

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1. Since the box moves with a constant speed, the acceleration is zero. According to Newton's second law, the net force must be zero. This means that the pulling force is exactly balanced by the frictional force.

2. The work done by a force is given by
   $$
   W = \vec{F} \cdot \vec{d} = Fd\cos\theta,
   $$
   where $$ \theta $$ is the angle between the force and the displacement. Since the forces along the horizontal direction cancel each other, the work done by the pulling force is positive and exactly canceled by the negative work done by friction.

3. The total (net) work done on the box is the sum of the work done by all forces. Therefore,
   $$
   W_{\text{total}} = W_{\text{applied}} + W_{\text{friction}} = 20\,\text{J} + (-20\,\text{J}) = 0\,\text{J}.
   $$

4. Hence, when the box covers a horizontal distance of 10 m, the total work done on the box is
   $$
   \boxed{0 \, \text{J}}.
   $$
The total work done on the box is 0 J.

4. A box of mass 0.50 kg is pulled by a force of magnitude \( F=2.0 \mathrm{~N} \) across a rough, horizontal surface as shown. If the box moves with constant speed along the straight path, what is the total work done on the box when it covers a horizontal distance of 10 m ?

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