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Anon25 [30]
2 years ago
6

If there is a negative sign in front of the Hooke's Law equation, what does the force, F, represent?

Physics
1 answer:
solong [7]2 years ago
3 0

Answer:

C) The restoring force

Explanation:

Hooke's Law states that the restoring force acting on a spring is given by the equation:

F=-kx

where

k is the spring constant

x is the displacement of the spring from its equilibrium position

The negative sign in the equation tells the direction of the restoring force. In fact, this force tends to bring the spring back to its equilibrium position: so, the force is always in opposite direction to the displacement.

This means that when the spring is stretched to the right, the restoring force tends to bring it back to the left, to the equibrium position; if the spring is compressed to the left, the restoring force tends to bring it back to the right, to the equilibrium position.

So the correct option is

C) The restoring force

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A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
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Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

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2 years ago
Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large cont
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<h2>For Second Solid Lumped System is Applicabe</h2>

Explanation:

Considering heat transfer between two identical hot solid bodies and their environments -

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When light hits the boundary between two different materials, it can undergo when light hits the boundary between two different
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2 years ago
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A girl rolls a ball up an incline and allows it to return to her. For the angle ! and ball involved, the acceleration of the bal
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Answer:

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The total time required for the ball to return to the child’s hand is 3.2654 s.

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The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t  to reach the top.

Using

v = u + a×t

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t = 1.6327 s

Since, this is the same time the ball will come back. So,

<u>Total time to go and come back = 2* 1.6327 = 3.2654 s </u>

To find the distance, using:

v² = u² + 2×a×s

0² = 4² + 2×(-2.45)×s

s = 3.2653 m

<u>Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.</u>

5 0
2 years ago
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