Rachel lifts a 7.00kg box onto a shelf 1.5m off of the floor. The box slips off of the shelf and falls to the floor. What is the velocity of the box the moment it impacts the floor (in meters per second)?



Answer :

Answer:

To determine the velocity of the box the moment it impacts the floor, we can use the conservation of mechanical energy. The potential energy of the box at the height of the shelf is converted into kinetic energy when it hits the floor.

The potential energy (PE) of the box at the height of the shelf can be calculated using the formula:

PE = m × g × h

Where:

m = mass of the box = 7.00 kg

g = acceleration due to gravity = 9.8 m/s^2 (on the surface of the Earth)

h = height from which the box falls = 1.5 m

Now, we can calculate the potential energy:

PE = (7.00 kg) × (9.8 m/s^2) × (1.5 m)

PE = 103.95 J

Since the box loses all of its potential energy as it hits the floor, the kinetic energy (KE) of the box just before it hits the floor is equal to the potential energy:

KE = PE

Now, we can calculate the kinetic energy:

KE = (7.00 kg) × (9.8 m/s^2) × (1.5 m)

KE = 103.95 J

To find the velocity of the box the moment it impacts the floor, we can use the formula:

KE = 0.5 × m × v^2

Where:

v = velocity of the box the moment it impacts the floor (in meters per second)

Rearranging the formula to solve for v, we get:

v = √(2 × KE / m)

Now, we can plug in the values:

v = √(2 × 103.95 J / 7.00 kg)

v = √(29.35 m^2/s^2)

v = 5.42 m/s

Therefore, the velocity of the box the moment it impacts the floor is approximately 5.42 meters per second.

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