A 60.0 kg person bends his knees and then jumps straight up. After his feet leave the floor, his motion is unaffected by air resistance and his center of mass rises by a maximum of 14.9 cm. Model the floor as completely solid and motionless.
a)With what momentum does the person leave the floor? kg-m/s up
v=(2x9.8x14.9)^(1/2)
p=60v
b)With what kinetic energy does the person leave the floor?
k=1/2(60)v^2
Your method is fine except the height H must be converted to meters. Also: why don't you compute v and provide numerical answers, with units (kg m/s and Joules)?
v = 1.7 m/s
To calculate the momentum with which the person leaves the floor, we need to find the velocity (v) with which the person jumps.
To find the velocity, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to the force of gravity acting on the person:
v^2 = u^2 + 2as
where:
v = final velocity (which is what we want to find)
u = initial velocity (which is 0 because the person starts from rest)
a = acceleration due to gravity (approximately 9.8 m/s^2)
s = displacement (which is the maximum height reached by the person, 14.9 cm or 0.149 m in this case)
Plugging in the values:
v^2 = 0 + 2 * 9.8 * 0.149
v^2 = 2 * 9.8 * 0.149
v^2 = 2.9232
v = √(2.9232)
v ≈ 1.71 m/s
Now that we have the velocity (v), we can calculate the momentum (p):
p = mass * velocity
p = 60 kg * 1.71 m/s
p ≈ 102.6 kg·m/s (upwards)
Therefore, the person leaves the floor with approximately 102.6 kg·m/s momentum upwards.
To calculate the kinetic energy with which the person leaves the floor, we can use the equation:
kinetic energy (KE) = 1/2 * mass * velocity^2
Plugging in the values:
KE = 1/2 * 60 kg * (1.71 m/s)^2
KE ≈ 87.8 joules
Therefore, the person leaves the floor with approximately 87.8 joules of kinetic energy.
a) To find the momentum with which the person leaves the floor, we first need to calculate the velocity at which the person jumps. We can use the equation v = (2gh)^(1/2), where v is the velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height or displacement (14.9 cm = 0.149 m). Plugging in the values, we have:
v = (2 x 9.8 x 0.149)^(1/2)
v ≈ 1.935 m/s
Next, we can calculate the momentum using the formula p = mv, where m is the mass of the person (60.0 kg):
p = 60.0 kg x 1.935 m/s
p ≈ 116.1 kg-m/s (upwards)
So, with what momentum the person leaves the floor is approximately 116.1 kg-m/s (upwards).
b) To find the kinetic energy with which the person leaves the floor, we can use the equation k = (1/2)mv^2, where k is the kinetic energy, m is the mass of the person (60.0 kg), and v is the velocity we calculated in part a (1.935 m/s):
k = (1/2) x 60.0 kg x (1.935 m/s)^2
k ≈ 111.9 J (joules)
So, the person leaves the floor with approximately 111.9 J of kinetic energy.