Can someone please help me with this one
1 answer:
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Answer:
a)
840 N
b)
10920 J
c)
- 10192 J
d)
4.3 m/s
Explanation:
a)
T = tension force in the cable in upward direction = ?
a = acceleration of the person in upward direction = 0.70 m/s²
m = mass of the person being lifted = 80 kg
Force equation for the motion of person in upward direction is given as
T - mg = ma
T = m (g + a)
T = (80) (9.8 + 0.70)
T = 840 N
b)
d = distance traveled in upward direction = 13 m
= Work done by tension force
Work done by tension force is given as
= T d
= (840) (13)
= 10920 J
c)
d = distance traveled in upward direction = 13 m
= Work done by person's weight
Work done by person's weight is given as
= - mg d
= - (80 x 9.8) (13)
= - 10192 J
d)
= Net force on the person = ma = 80 x 0.70 = 56 N
v₀ = initial speed of the person = 0 m/s
v = final speed
Using work-energy theorem
d = (0.5) m (v² - v₀²)
(56) (13) = (0.5) (80) (v² - 0²)
v = 4.3 m/s
To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is
Where,
F = Force
r = Distance
Replacing,
The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration
I = Moment of inertia
= Angular acceleration
Replacing with our values we have that
Therefore the value of moment of inertia is