Question

В шахту равноускоренно опускается бадья массой 280кг. В первые 10 сек она проходит расстояние, равное 35м. Найдите силу натяжения каната, к которому подвешена бадья.

Answer 1

По 2-му закону Ньютона в проекции на ось OY имеем: T-mg=-ma
где Т - сила натяжения
T=mg-ma=m(g-a)
S=a t^2/2 ---->a=2*S/t^2 =2*35/10^2=70/100=0,7  м/c^2
T=280*(9,8-0,7)=2548 H

Answer 2

The Question

The question asks to find the tension force in the rope suspending a bucket that is lowered down a mine shaft with constant acceleration.

The given details are: - Mass of the bucket (m) = 280 kg - Distance covered by the bucket in the first 10 seconds (s) = 35 m

Solving the Question

We are given that the bucket is lowered down the mine shaft with constant (equal) acceleration. When an object is undergoing uniform acceleration, we can use the kinematic equations of motion to relate its displacement (s), time (t), velocity (v), and acceleration (a).

For uniform acceleration, the kinematic equation relating displacement, time, initial velocity (which is 0 in this case) and acceleration is:

`s = ut + 1/2 * a * t^2`

We know: s = 35 m (distance covered in 10 s) t = 10 s u = 0 m/s (initial velocity is 0)

Plugging these values in the equation: 35 = 1/2 * a * 10^2 a = 7 m/s^2

We can use Newton's Second Law, `F = m * a`, to find the tension force F in the rope. Here: m = 280 kg a = 7 m/s^2

Plugging in the values: F = m * a = 280 * 7 = 1960 N