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 be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.
First, we manipulate the one of the kinematic equations
v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped
Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,
<span>we get v0 = sqrt (2(a)(x))
Substituting the known values,
v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
</span>
Therefore, before stopping the car the original speed of the car would be 14.49 m/s
The equation formula:
P V = n R T
1,245 * 2 l = n R * 300 K
n R = 1,245 * 2 : 300 = 8.3
P * 2.5 l = n R * 400 K
P * 2.5 = 8.3 * 400
P = 3,320 : 2.5 = 1328 J
Answer: A ) 1,328 joules.
The mass of the rider and bicycle would be 35.7 and I’m assuming the unit would be kg so 35.7kg
Every action has an equal or opposite reaction.
You weigh 60kg
<span>So your acceleration is 6N / 60 kg = 0.1m/s^2</span>
