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Gnoma [55]
3 years ago
8

. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat

ion down the ramp and the force of the ramp on the block? (b) What force applied upward along and parallel to the ramp would allow the block to move with constant velocity?
Physics
1 answer:
AfilCa [17]3 years ago
5 0

Answer:

a) a = 4.9 m / s²,  N = 16.97 N   and b)   F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

        Wy = W cos 30

Let's write the equations on each axis

X axis

        Wx = ma

Y Axis  

       N- Wy = 0

       N = Wy = mg cos 30

       N = 2.0 9.8 cos 30

       N = 16.97 N

We calculate the acceleration

       a = Wx / m

       a = mg sin 30 / m

       a = g sin 30

       a =9.8 sin 30

       a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N

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<h2>QUESTION:- It is easier to lift the same load by using three pulley system than by using two-pulley system.</h2>

<h2>ANSWER:- IN CASE OF IDEAL PULLEY SYSTEM</h2>

<h2>REASON:- </h2>

Logic behind is lies behind the mechanical advantage of the provided bt the Pulley system.

as if we calculate the mechanical advantage of the 2 Pulley system we will have the value 2

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Assume that the body's muscle mechanism can be approximated by a spring with a uniform continuous mass distribution that follows
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Based on Hooke's law, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

<h3>What is the spring constant?</h3>

The spring constant or stiffness constant of an elastic spring is constant which describes the extent a bit forceapplied to an elastic spring will extend it.

  • Spring constant, K = force/extension

Assuming, a body's muscle mechanism is a spring obeying Hooke's law, the effective mass of the spring with mass m is 1/3 of the mass of the spring = m/3

The potential energy that can be stored = ke^2 / 2

where K is spring constant and e is the extension produced.

Therefore, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

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At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti
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Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

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Substituting

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