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Ira Lisetskai [31]
3 years ago
13

g A uniform ladder whose length is 4.4 m and whose weight is 330 N leans against a frictionless vertical wall. The coefficient o

f static friction between the level ground and the foot of the ladder is 0.46. What is the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping

Physics
1 answer:
Damm [24]3 years ago
3 0

Answer:

Distance = 4.4 [m]

Explanation:

This problem can be easily solved using a static analysis of forces acting on the ladder, taking into account the respective distances. For easy understanding, a free body diagram should be made.

We perform a sum of force on the X-axis equal to zero, to find that the force exerted by the wall is equal to the friction force on the floor.

Then we perform a summation of forces on the Y axis, to determine that the normal force exerted by the floor is equal to the weight of the ladder.

We know that the friction force is equal to the product of normal force by the coefficient of friction.

In this way, by relating the friction force to the equations deduced above we can find the force exerted by the wall.

Then we make a summation of moments around the base point of the ladder, the equation realized can be seen in the attached image.

In the last analysis we can find the relationship between the horizontal and vertical distance of the ladder, with respect to the wall and the floor.

Then with the complementary analysis of the Pythagorean theorem we can find another additional equation.

The result of the greater distance is 4.4 [m]

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Answer:

The question does not state how the answer is to be entered. I would use 10% because that is most common.

Explanation:

You are using 300 J of energy to get 30 J of light energy. The fact that you can account for the left over energy is not relevant.

Efficiency = Work Out / Work In

Efficiency = 30 J / 300 J  = 0.1

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Explanation:

Does this help ?

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3 years ago
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Lars is standing near the edge of a 90-meter cliff. He throws a ball upward, but does not catch it, and it falls to the bottom o
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Answer:

c. 43 m/s

Explanation:

Given the following data;

Displacement, S = 90 meters

Time, t = 5.55 seconds

To find the initial velocity;

We would use the second equation of motion given by the formula;

S = ut + \frac {1}{2}at^{2}

Where;

  • S represents the displacement or height measured in meters.
  • u represents the initial velocity measured in meters per seconds.
  • t represents the time measured in seconds.
  • a represents acceleration measured in meters per seconds square.

We know that acceleration due to gravity is -9.8m/s² because the direction is downward.

Substituting into the equation, we have;

90 = u*5.55  + \frac {1}{2}*(-9.8)*5.55^{2}

90 = u5.55 - 4.9*30.8025

90 = u5.55 - 150.93225

Rearranging the equation, we have;

u5.55 = 90 + 150.93225

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