What is the weight of a 300kg human on earth
1 answer:
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(a)
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:
where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,
And the average force on the person is given by
with m = 75.0 kg being the mass of the person. Substituting,
where the negative sign means the force is opposite to the direction of motion of the person.
b)
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is
So the average force on the person is
The question is incomplete. The complete question is :
In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?
Solution :
Let M.I. of disk A =
So, M.I. of disk B =
Angular velocity of A =
So the kinetic energy of the disk A =
After coupling, the angular velocity of both the disks will be equal to ω.
Angular momentum will be conserved.
So,
Now,
Therefore, the maximum initial K.E. = 3066.67 J
