A boy weighing 3.9 102 N jumps from a height of 2.15 m to the ground below. Assume that the force of the ground on his feet is constant.
a) Compute the force of the ground on his feet if he jumps stiff-legged, the ground compresses 1.80 cm, and the compression of tissue and bones is negligible.
b) Compute the force his legs exert on his upper body (trunk, arms, and head), which weighs 2.50 102 N, under the conditions assumed above.
c) Now suppose that his knees bend on impact, so that his trunk moves downward 45.0 cm during deceleration. Compute the force his legs exert on his upper body.
To find the answers to all three parts of the problem, we need to apply the principles of physics, specifically Newton's laws of motion. Let's break down each part step by step.
a) To compute the force of the ground on the boy's feet when he jumps stiff-legged, we need to consider the concept of work and energy. When the boy jumps and lands, his potential energy is converted into kinetic energy. Therefore, we can equate the potential energy at the initial height to the kinetic energy at the compressed height.
The potential energy at a height of 2.15 m can be calculated using the formula: Potential energy = mass * gravity * height.
Given that mass = 3.9 * 10^2 N and gravity = 9.8 m/s^2, the potential energy is:
Potential energy = (3.9 * 10^2 N) * (9.8 m/s^2) * (2.15 m) = 8373 J.
Now, the compression of the ground can be considered as work done on the boy's body. We'll use the formula: Work = force * distance * cos(theta), where theta is the angle between the force and displacement, which is 0 degrees here because the force is vertical.
Let's find the distance first. The compression of the ground is given as 1.80 cm, which is equivalent to 0.018 m.
Now, we can rearrange the formula for work to solve for force:
Force = Work / (distance * cos(theta)).
Since cos(theta) = cos(0) = 1, the formula for force reduces to:
Force = Work / distance.
Substituting the values, Force = 8373 J / 0.018 m = 465,166.67 N.
Therefore, the force of the ground on his feet when he jumps stiff-legged is approximately 465,166.67 N.
b) To compute the force his legs exert on his upper body (trunk, arms, and head), we need to apply Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
The force of the ground on his feet from part (a) will be balanced by an equal and opposite force exerted by his legs on his upper body. Therefore, the force his legs exert on his upper body will also be approximately 465,166.67 N.
c) In this scenario, where the boy's knees bend on impact, we need to consider the concept of impulse. Impulse is defined as the change in momentum and is equal to the force applied multiplied by the time of contact.
We can calculate the impulse using the formula: Impulse = force * time.
Given that the trunk moves downward 45.0 cm during deceleration, which is equivalent to 0.45 m, and assuming the deceleration is uniform, we can calculate the time using the formula: time = distance / velocity.
The velocity can be calculated using the formula: Final velocity = Initial velocity + (2 * acceleration * distance).
The initial velocity can be assumed to be zero since the boy is coming to a stop. Therefore, the formula for velocity becomes: Final velocity = 2 * acceleration * distance.
Given that acceleration = (final velocity - initial velocity) / time, and rearranging the equation to solve for time, we have: time = distance / ((final velocity - initial velocity) / 2).
Plugging in the values, time = 0.45 m / ((0 - 0) / 2) = 0.45 m / 0 = undefined.
Since time is undefined, it means that the deceleration is instantaneous, assuming no external forces acting on the boy's body. Therefore, the force his legs exert on his upper body can be considered infinite or extremely large.
Please note that in actual scenarios, the deceleration process takes a finite amount of time, and there are various factors like shock absorption by tissues and clothes that affect the force exerted by the legs on the upper body.