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A 7 kg box is drpooed from the third floor (10 meters hight) to the ground. a. What is the maximum potential energy associated with the box PE = mgh = 7(9.8) (10) = 686 J b. Ignoring friction, find thefind the box's speed just before it hits the ground. KE; + PE: = KEp + PEg 0 + PE; = KE,+0 1 mgh = -mv2 2 ひ = 2mgh = 14 m/s



Answer :

audreyn06

Answer: 686 J; 14 m/s

Explanation:

The gravitational potential energy of an object is given by the formula:

U = mgh, where:

  • U is the gravitational potential energy
  • m is the mass of the object
  • g is the acceleration due to gravity (we will use 9.8 m/s²)
  • h is the height above the ground

We are given:

  • m = 7 kg
  • h = 10 m

Let's plug the given values into the formula:

U = 7·9.8·10 = 686 J

To find the box's speed just before it hits the ground, we can apply the law of conservation of energy, which states that energy cannot be created nor destroyed, only transferred. This means that the maximum gravitational potential energy has to equal the maximum kinetic energy, which the box reaches just before it hits the ground. The formula for kinetic energy is given as:

K = (1/2)mv², where:

  • K is the kinetic energy
  • m is the mass
  • v is the velocity

Since the maximum gravitational potential energy is equivalent to the maximum kinetic energy of the box:

U = K

686 = (1/2)mv²

686 = (1/2)7v²

Solve for v:

1372 = 7v²

v² = 196

v = 14 m/s

The maximum potential energy associated with the box is 686 J and the box's speed just before it hits the ground is 14 m/s.

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