Question

a light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. if the block is released from rest at a distance of 2.04 m above the floor, what is its speed just before it strikes the floor?

Answer 1

To calculate the block's speed just before it strikes the floor, the principle of conservation of energy is applied; converting the initial potential energy to kinetic energy gives a speed of about 6.32 m/s.

Step-by-step explanation:

To find the speed of the 1.50-kg block just before it strikes the floor, we apply the principle of conservation of energy. Since the block is released from rest, all its initial potential energy (gravitational) will be converted into kinetic energy just before it hits the floor.

The initial potential energy (PE) is given by the formula:

• PE = m * g * h

where m is the mass of the block, g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height from which the block was released.

The kinetic energy (KE) just before impact will be the same as the initial potential energy (assuming no energy is lost to friction or air resistance). The kinetic energy is given by:

• KE = 1/2 * m * v2

Setting the initial potential energy equal to the kinetic energy, we have:

• m * g * h = 1/2 * m * v2

Cancelling out the mass (m) from both sides and solving for v (speed), we get:

• v = sqrt(2 * g * h)

Plugging in the known values, we get the speed just before the block strikes the floor:

• v = sqrt(2 * 9.81 m/s2 * 2.04 m)
• v ≈ 6.32 m/s

Therefore, the speed of the block just before impact is approximately 6.32 m/s.