F=150Н; x=100+5t+0,5t^2

Предыдущий
вопрос Следующий
вопрос

To find the work done by the force F over the distance x, we can integrate the force function with respect to x:

W = ∫ F dx

Given F = 150N and x = 100 + 5t + 0.5t^2, we need to first express dx in terms of dt:

dx/dt = d(100 + 5t + 0.5t^2)/dt
= d(100)/dt + 5 + 0.5(2t)
= 5 + t

Now we can express dx in terms of dt:

dx = (5 + t) dt

Substitute dx into the work equation:

W = ∫ F dx
= ∫ 150 dx
= ∫ 150(5 + t) dt
= 150∫ (5 + t) dt
= 150(5t + 0.5t^2) + C
= 750t + 75t^2 + C

Therefore, the work done by the force F over the distance x is 750t + 75t^2 + C.