Suppose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 51.5 N.

If he does 3.85 102 J of work on the sled while exerting the force horizontally, through what distance must he have pushed it?

Just for clarification, the question asks if he does 3.85 X 10^2 J of work.

What force is needed to accelerate a child on a sled (total mass = 60 kg) at 1.25 m/s2?

To find the distance through which the Eskimo must have pushed the sled, we can use the formula for work:

Work = Force × Distance

Given that the Eskimo exerts a force of 51.5 N and does 3.85 × 10^2 J of work, we can rearrange the formula to solve for distance:

Distance = Work / Force

Plugging in the values from the problem:

Distance = (3.85 × 10^2 J) / (51.5 N)

Calculating this:

Distance = 7.4776 meters

Therefore, the Eskimo must have pushed the sled through a distance of approximately 7.48 meters.

To find the distance the Eskimo must have pushed the sled, we can start by using the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy.

The work done by the Eskimo on the sled can be calculated using the formula:

Work = Force × Distance × cos(θ)

Where:
Work is the energy transferred to the object, in joules (J)
Force is the applied force on the object, in newtons (N)
Distance is the displacement of the object, in meters (m)
θ is the angle between the force and the direction of displacement

In this case, the force and displacement are horizontal, so the angle θ is 0 degrees (cos(0) = 1).

We are given:
Force = 51.5 N (applied force by the Eskimo)
Work = 3.85 × 10^2 J

Rearranging the formula, we have:

Distance = Work / (Force × cos(θ))

Substituting the given values:

Distance = (3.85 × 10^2 J) / (51.5 N × cos(0))

Since cos(0) = 1, the equation simplifies to:

Distance = (3.85 × 10^2 J) / 51.5 N

Calculating the value, we get:

Distance ≈ 7.48 meters (rounded to two decimal places)

Therefore, the Eskimo must have pushed the sled for approximately 7.48 meters.