An 82 kg man lowers himself to the ground from a height of 6.6 m by holding onto a rope that runs over a frictionless pulley to a 66 kg sandbag. With what speed does the man hit the ground if he started from rest?
Force up on 66 kg mass = T
Force down on 66 kg mass = 66(9.8 )
net force up on 66 kg = T-66(9.8)
F = m a
T -66(9.8) = 66 a
a is down
Force down on 82 kg mass = 82(9.8)
Force up on 82 kg mass = T
net force down on 82 kg = 82(9.8) -T
F = m a
82(9.8) - T = 82 a
combine by addition
(82-66)(9.8) = (82+66) a
solve for a which is down
then
6.6 = (1/2) a t^2
solve for t
then v = a t
To find the speed at which the man hits the ground, we can use the principle of conservation of mechanical energy.
The potential energy of the man at the top is given by the equation:
PE = m * g * h
Where:
PE is the potential energy,
m is the mass of the man (82 kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
h is the height (6.6 m).
Substituting the values, we get:
PE = 82 kg * 9.8 m/s^2 * 6.6 m
Now, the potential energy is converted into kinetic energy when the man reaches the ground. The kinetic energy is given by the equation:
KE = 1/2 * m * v^2
Where:
KE is the kinetic energy,
m is the mass of the man (82 kg),
v is the velocity of the man.
Since the man starts from rest, his initial kinetic energy is 0. Therefore, we can equate the potential energy to the kinetic energy:
PE = KE
82 kg * 9.8 m/s^2 * 6.6 m = 1/2 * 82 kg * v^2
Simplifying the equation:
(v^2)/2 = 9.8 m/s^2 * 6.6 m
v^2 = 2 * 9.8 m/s^2 * 6.6 m
v^2 = 128.52 m^2/s^2
Taking the square root of both sides of the equation:
v = √(128.52 m^2/s^2)
v ≈ 11.342 m/s
Therefore, the man hits the ground with a speed of approximately 11.342 m/s.