A wood block falls from a shelf and hits the ground after 1.2 s. The mass of the block is .25kg. The block hits the floor and comes to rest in .1s. Determine the force applied by the floor

It hits the floor with velocity of

V = g*1.2s = 11.8 m/s

Use that and the mass to get the momentum, M*V = 2.94 kg m/s

The average force applied by the floor during deceleration (t' = 0.1 s) is M*V/t' = ___ Newtons

(impulse = F*t' = momentum change)

To determine the force applied by the floor, we need to use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

Given:
Mass of the block (m) = 0.25 kg
Time taken to come to rest (t) = 0.1 s

First, let's calculate the acceleration (a) of the block using the equation:

a = (final velocity - initial velocity) / time

Since the block comes to rest, the final velocity (v) is 0, and the initial velocity (u) can be calculated from the time it takes to fall using the equation:
u = g * t

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

u = 9.8 m/s^2 * 1.2 s = 11.76 m/s

Now, we can calculate the acceleration:

a = (0 m/s - 11.76 m/s) / 0.1 s
= -117.6 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity (which is upward).

Finally, we can find the force (F) using Newton's second law:

F = m * a
= 0.25 kg * (-117.6 m/s^2)
≈ -29.4 N

The negative sign implies that the force is directed upwards, opposing the weight of the block. If we take the magnitude of the force, it will be 29.4 N.

Therefore, the force applied by the floor is approximately 29.4 N upwards.