In order for work to take place
1 answer:
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Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Answer:
Speed of the airplane 10.0 s later = 12.2 m/s
Explanation:
Mass of Boeing 777 aircraft = 300,000 kg
Braking force = 445,000 N
Deceleration
Initial velocity, u = 27 m/s
Time , t = 10 s
We have equation of motion, v =u +at
v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s
Speed of the airplane 10.0 s later = 12.2 m/s