To solve this problem, we can use the work-energy principle. The work-energy principle states that the work done on an object is equal to its change in kinetic energy.
The work done by the gravitational force as the body falls through a height of 40 m is given by:
Work_gravity = mgh
= (1/4 kg)(9.8 m/s^2)(40 m)
= 98 J
The work done by the sand in bringing the body to rest is equal to the work done by the average force exerted by the sand multiplied by the distance the body penetrates into the ground. We can calculate this work using the formula:
Work_sand = Force_sand x distance_penetrated
To find the average force exerted by the sand, we need to know the final velocity of the body just before it comes to rest. We can find this using the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (0 m/s, since the body comes to rest)
vi = initial velocity (unknown)
a = acceleration (acceleration due to gravity = 9.8 m/s^2)
d = distance traveled (1/2 m into the sandy ground)
Solving for vi, we get:
vi^2 = vf^2 - 2ad
= (0 m/s)^2 - 2(9.8 m/s^2)(-1/2 m)
= 4.9 m^2/s^2
Taking the square root of both sides, we find:
vi = 2.2 m/s
Now, using the work-energy principle:
Work_sand = Work_gravity
Force_sand x distance_penetrated = mgh
Force_sand = (mgh) / distance_penetrated
= (1/4 kg)(9.8 m/s^2)(40 m) / (1/2 m)
= 98 J / (1/2 m)
= 196 N
Therefore, the average force exerted by the sand in bringing the body to rest is 196 Newtons.
A body of mass 1/4 kg falls from rest through a height of 40 m and comes to rest having penetrated a distance of 1/2 m into the sandy ground . Calculate the average force exerted by the sand in bringing the body to rest
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1 answer