Work Done by Force on Balky Cow: -209J

[Heat]

[SOLVED] Balky Cow (Forces)

Homework Statement

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component F_x = -[20.0N + (3.0N/m x].

How much work does the force you apply do on the cow during this displacement?

The Attempt at a Solution

Given that: F_x = -[20.0N + (3.0N/m x]

Displacement of 6.9

F_x = -[20.0N + (3.0N/m (6.9)] = -40.7N
F_xo = -[20.0N + (3.0N/m (0)] = -20 N

w= k2-k1
k = 1/2mv^2
W = F*s

W = (-40.7-20) * (6.9m)
= 418.83 :(

Solution is: -209J why?

 

[learningphysics]

$$Work = \int_0^{6.9} F_x dx$$

you need to calculate this integral...

you can calculate the integral... or you can plot the Fx vs. x graph... and the area under this graph is the work...