Ермек жүк түсіріп жүретін адамның 60 Н-нан аз мөлшерде күш түсіредігін анықтады. Кораптың бір қорағын жылжытуға

Шешім:

A person carrying a load experiences a force of 60 N in the direction of motion. Let"s assume this force is \(F_{\text{{load}}}\).

The force required to move the load can be found using the equation for force:

\[F = m \cdot a\]

Where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration.

We can rearrange this equation to solve for acceleration:

\[a = \frac{F}{m}\]

Now, we need to determine the mass of the load. The weight of an object can be calculated using the equation:

\[F = m \cdot g\]

Where \(F\) is the force (weight), \(m\) is the mass, and \(g\) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)).

Rearranging this equation, we can solve for mass:

\[m = \frac{F}{g}\]

Substituting the given force of 60 N and the acceleration due to gravity of 9.8 m/s\(^2\), we can calculate the mass:

\[m = \frac{60 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 6.12 \, \text{kg}\]

Now that we have the mass, we can calculate the acceleration using the first equation:

\[a = \frac{F_{\text{load}}}{m} = \frac{60 \, \text{N}}{6.12 \, \text{kg}} \approx 9.80 \, \text{m/s}^2\]

So, the load is experiencing an acceleration of approximately 9.80 m/s\(^2\).

Now, let"s consider the force required to drag the load along the incline. We can analyze the forces acting on the load in the vertical and horizontal directions.

In the vertical direction, we have the force of gravity pulling the load downward and the normal force exerted by the inclined plane pushing upward. Since the load is being dragged horizontally, there is no vertical acceleration, and these two forces must balance each other:

\[F_{\text{gravity}} = F_{\text{normal}}\]

The force of gravity can be calculated using:

\[F_{\text{gravity}} = m \cdot g\]

Substituting the mass we calculated earlier, we have:

\[F_{\text{gravity}} = 6.12 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \approx 59.98 \, \text{N}\]

Therefore, the normal force exerted by the inclined plane is approximately 59.98 N.

In the horizontal direction, the force required to drag the load is equal to the force of friction between the load and the inclined plane. This force can be calculated using the equation:

\[f = \mu \cdot F_{\text{normal}}\]

Where \(f\) is the force of friction and \(\mu\) is the coefficient of friction between the load and the inclined plane.

Substituting the given value of the coefficient of friction of 0.3 and the normal force we calculated earlier, we can find the force of friction:

\[f = 0.3 \cdot 59.98 \, \text{N} \approx 17.99 \, \text{N}\]

So, the force required to drag the load along the incline is approximately 17.99 N.