Question:

A 5 kg block is initially at rest at the edge of a table that is 1.2 m above the ground. The table has a coefficient of kinetic friction of 0.3. The block is pushed horizontally with a force of 10 N, causing it to slide off the table and fall to the ground. Assume acceleration due to gravity as 9.8 m/s^2.

a) Determine the work done by the frictional force as the block slides off the table. b) Calculate the speed of the block just before it hits the ground. c) Determine the total mechanical energy of the block just before it hits the ground.

Answer:

a) To determine the work done by the frictional force, we need to find the frictional force acting on the block. The frictional force can be calculated using the equation:

Frictional force (Ff) = coefficient of friction (μ) × normal force (N)

The normal force is equal to the weight of the block, which can be calculated as:

Normal force (N) = mass (m) × acceleration due to gravity (g)

Normal force (N) = 5 kg × 9.8 m/s^2 = 49 N

Now, we can calculate the frictional force:

Frictional force (Ff) = 0.3 × 49 N = 14.7 N

The work done by the frictional force can be calculated using the equation:

Work (W) = force (F) × displacement (s)

Since the force of friction is the only force doing work horizontally, the displacement is the horizontal distance the block travels before leaving the table. The horizontal distance can be calculated using the equation:

Horizontal distance (s) = height of the table (h)

Horizontal distance (s) = 1.2 m

Therefore, the work done by the frictional force is:

Work (W) = 14.7 N × 1.2 m = 17.64 J

b) To calculate the speed of the block just before it hits the ground, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the block is equal to its final mechanical energy.

The initial mechanical energy includes the gravitational potential energy and the initial kinetic energy, which is zero since the block is initially at rest.

The final mechanical energy includes the final gravitational potential energy and the final kinetic energy. At the ground level, the gravitational potential energy is zero, so only the final kinetic energy remains.

Since the mechanical energy is conserved, we can write the equation:

Initial mechanical energy = Final mechanical energy

mgh + 0 = 0.5mv²

Where: m = mass of the block = 5 kg g = acceleration due to gravity = 9.8 m/s^2 h = height of the table = 1.2 m v = final velocity

Therefore, we have:

5 kg × 9.8 m/s^2 × 1.2 m = 0.5 × 5 kg × v²

58.8 J = 2.5v²

Dividing both sides by 2.5:

v² = 23.52 m²/s²

Taking the square root of both sides:

v ≈ 4.85 m/s

Therefore, the speed of the block just before it hits the ground is approximately 4.85 m/s.

c) The total mechanical energy of the block just before it hits the ground can be calculated using the final kinetic energy:

Total mechanical energy = Final kinetic energy = 0.5mv²

Total mechanical energy = 0.5 × 5 kg × (4.85 m/s)²

Total mechanical energy ≈ 58.96 J

Therefore, the total mechanical energy of the block just before it hits the ground is approximately 58.96 J.