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...

 

[ogb p]

The work done by a force is calculated by multiplying the force by the displacement in the direction of the force. In this case, the force applied by the scientist is in the negative x-direction, while the cow's displacement is in the positive x-direction. This means that the work done by the force is negative, indicating that the force is opposing the motion of the cow.

To calculate the work done, we use the equation W = F * d. In this case, the force is given by F_x = -[20.0N + (3.0N/m x], and the displacement is 6.9m. Plugging in these values, we get:

W = (-40.7N) * (6.9m) = -280.83J

However, this value is not the final answer. The problem states that the cow walks from x = 0 to x = 6.9m, but the force is only applied from x = 0 to x = 6.9m. This means that the cow has already traveled 6.9m before the force is applied, and this distance does not contribute to the work done by the force. Therefore, we must subtract this distance from the total displacement, giving us a final displacement of 6.9m - 6.9m = 0m. Plugging this into the work equation again, we get:

W = (-40.7N) * (0m) = 0J

However, this still is not the final answer. The problem also states that the force applied is given by F_x = -[20.0N + (3.0N/m x]. This means that the force is not constant throughout the displacement, but rather changes as the cow moves. In order to accurately calculate the work done, we must use calculus to integrate the force function over the displacement. This gives us:

W = ∫ F_x dx from x = 0 to x = 6.9

= ∫ (-20N - 3N/m * x) dx from x = 0 to x = 6.9

= [-20x - (3/2) * x^2] from x = 0 to x = 6.9

= (-20 * 6.9) - (3/2 * 6.9^2) - (-20 * 0) -