A Stone of 5 kg falls from the top of a cliff 50 m high and buries 1m in sand. Find the average resistance offered by the way sand and the time taken to penetrate.
force*distance=energyfromfalling= mgh
force=5*9.8*50/5 N
time:
force*time= mass*changeinvelocity
change in velocity:
1/2 m vf^2=mgh
vf=sqrt(2gh)
time= mass*sqrt(2gh)/forceabove
To find the average resistance offered by the sand and the time taken to penetrate, we need to use equations of motion and apply Newton's laws of motion.
First, we'll calculate the time taken by the stone to fall from the top of the cliff to the sand using the equation:
h = (1/2) * g * t^2
Here, h is the height of the cliff (50 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.
Rearranging the equation to solve for t:
t^2 = (2h) / g
Plugging in the values:
t^2 = (2 * 50) / 9.8
t^2 = 10.20
t = sqrt(10.20)
t ≈ 3.19 seconds
So, it takes approximately 3.19 seconds for the stone to fall from the top of the cliff to the sand.
Next, we'll calculate the average resistance offered by the sand. The resistance can be calculated using Newton's second law of motion:
F = m * a
Where F is the force, m is the mass, and a is the acceleration.
In this case, the force is equal to the weight of the stone, which can be calculated using:
F = m * g
F = 5 kg * 9.8 m/s^2
F = 49 N
Since the stone buries 1 m into the sand, the acceleration can be calculated using:
a = (v^2 - u^2) / (2 * s)
Where v is the final velocity, u is the initial velocity (which is 0 in this case since the stone starts from rest), and s is the displacement.
Here, v = √(2 * g * s)
v = √(2 * 9.8 * 1)
v ≈ √19.6
v ≈ 4.43 m/s
So, the acceleration is:
a = (4.43^2 - 0) / (2 * 1)
a = (19.60 - 0) / 2
a = 9.80 m/s^2
Finally, we can calculate the average resistance offered by the sand using:
F = m * a
R = F / m
R = (49 N) / 5 kg
R = 9.8 N
Therefore, the average resistance offered by the sand is 9.8 N.