To calculate the work done by the student on the desk, we need to consider the work done against friction since the desk is moving at a constant speed. The formula for work is:

$W = F_{\text{friction}} \cdot d \cdot \cos(\theta)$

where:

• $W$ is the work done,
• $F_{\text{friction}}$ is the force of kinetic friction,
• $d$ is the distance over which the desk is moved, and
• $\theta$ is the angle between the force and the direction of motion.

Since the desk is moving horizontally, the force exerted by the student is in the same direction as the motion, so $\theta = 0^\circ$, and $\cos(0^\circ) = 1$. Therefore, the work equation simplifies to:

$W = F_{\text{friction}} \cdot d$

Now, we can calculate the force of kinetic friction:

$F_{\text{friction}} = \mu_k \cdot N$

where:

• $\mu_k = 0.461$ is the coefficient of kinetic friction,
• $N$ is the normal force, which for a horizontal surface is equal to the weight of the desk, i.e., $N = mg$.

Given:

• $m = 71 \, \text{kg}$ (mass of the desk),
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity),
• $d = 6 \, \text{m}$ (distance the desk moves).

First, calculate the normal force:

$N = mg = 71 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 696.51 \, \text{N}$

Next, calculate the frictional force:

$F_{\text{friction}} = \mu_k \cdot N = 0.461 \times 696.51 \, \text{N} = 321.56 \, \text{N}$

Now, calculate the work done:

$W = F_{\text{friction}} \cdot d = 321.56 \, \text{N} \times 6 \, \text{m} = 1929.37 \, \text{J}$

Rounding to three significant figures:

$W = 1930 \, \text{J}$

So, the work done by the student on the desk is:

$W = 1930 \, \text{J}$

Let me know if you need further clarification or if you'd like more examples!

Related Questions:

1. How would the work change if the student pushed the desk with more force?
2. What happens to the work done if the coefficient of friction increases?
3. How would the work change if the student pushed the desk on an inclined surface?
4. How is the work done by friction related to energy?
5. What would happen to the work if the desk was pushed over a longer distance?

Tip:

When calculating work against friction, remember that the work done is always in the direction of the force applied, and friction always opposes the motion.