A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.
1 answer:
To solve this problem we will apply the concepts of equilibrium and Newton's second law.
According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So
Here,
T = Tension
m = Mass
g = Gravitational Acceleration
a = Acceleration (upward)
Rearranging to find T,
Therefore the tension force in the cable is 10290.15N
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Answer:
31.55 m/s
Explanation:
Let the initial velocity of the arrow is u metre per second.
Angle of projection, θ = 40 degree
range = 100 m
Use the formula for the range.
100 = u^2 Sin(2 x 40) / 9.8
100 x 9.8 = u^2 Sin 80
u^2 = 995.11
u = 31.55 m/s
Answer:
500 kg
Explanation:
It is given that,
The mass of a open train car, M = 5000 kg
Speed of open train car, V = 22 m/s
A few minutes later, the car’s speed is 20 m/s
We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :
initial momentum = final momentum
Let m is final mass
MV=mv
Water collected = After mass of train - before mass of train
= 5500 - 5000
= 500 kg
So, 500 kg of water has collected in the car.