A person jumps from the roof of a house 3.0 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.60 m. The mass of his torso (excluding legs) is 48 kg.
*How do I find the average force exerted on his legs by his torso during deceleration?
force*distance=mgh
workdoneinabsorbind enery=energyfromfall
solve for force.
but i solved for the force and it did not come out as the right anser i did Fd=mgh then said F(.6)=(48)(9.8)(3) but its wrong???
I don't know. Do you think it means on each leg individually?
hi can u help me in my home work
state the law of conservation of linear momentum and describe its consequences.
To find the average force exerted on the legs by the torso during deceleration, you can use Newton's second law of motion, which states that the force exerted on an object is equal to its mass multiplied by its acceleration. Here's how you can go about finding the average force:
1. First, let's calculate the initial velocity of the person before he lands on the ground. Since the person jumps vertically downwards, the vertical motion can be considered as free fall. The height of the house is given as 3.0 m, so we can use the equation for free fall displacement:
d = (1/2) * g * t^2
Where d is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time in seconds. Solving for t, we have:
t = sqrt((2 * d) / g)
t = sqrt((2 * 3.0) / 9.8)
t ≈ 0.782 s
2. Since the person bends their knees and decelerates over a distance of 0.60 m, we can calculate the deceleration using the equation for average acceleration:
a = (vf - vi) / t
where vf is the final velocity (0 m/s, as the person comes to rest) and vi is the initial velocity, which we'll calculate next.
vi = d / t
vi = 3.0 / 0.782
vi ≈ 3.84 m/s
a = (0 - 3.84) / 0.782
a ≈ -4.92 m/s^2 (negative sign indicates deceleration)
3. Now, we can use Newton's second law to find the force exerted on the legs by the torso. The mass of the torso is given as 48 kg, and the deceleration is -4.92 m/s^2:
F = m * a
F = 48 * -4.92
F ≈ -236.16 N
The negative sign indicates that the direction of the force exerted on the legs by the torso is opposite to the motion.
Therefore, the average force exerted on his legs by his torso during deceleration is approximately 236.16 N, in the opposite direction to the motion.