# Please see the link for question # 10-8) a,b,c ?? http://people.westminstercollege.edu/faculty/ccline/courses/phys211/hw-f07/hw-211-u10.pdf

Asked on by olkine

### 1 Answer

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

a) You should notice that the problem provides the force versus position graph, hence, you may evaluate the work done by the force calculating the area under the line.

Hence, evaluating the work done on the cart by the air pushing back on the fan as the cart moves from 5.0m to 0.0m, yields:

`W = DeltaF_1Deltax_1 + DeltaF_2Deltax_2 + DeltaF_3Deltax_3`

`Deltax_1 = 5.0m - 3.0m = 2.0m`

`Deltax_2 = 3.0m - 1.0m = 2.0m`

`Deltax_3 = 1.0m - 0.0 = 1.0m`

You need to us the information given by the graph to evaluate the forces `DeltaF_1,DeltaF_2,DeltaF_3` , such that:

`DeltaF_1 = 0.5N`

`DeltaF_2 =0.5N - 0.2N=0.3N`

`DeltaF_3 = 0.2N`

`W = 0.5N*2.0m + 0.3N*2.0m + 0.2N*1.0m`

`W = 1J + 0.6J + 0.2J => W = 1.8J`

Hence, evaluating the work done on the cart by the air pushing back on the fan as the cart moves from 5.0m to 0.0m, yields `W = 1.8J` .

b) You need to remember that the work done is equal to the change in kinetic energy of the object, such that:

`DeltaE_k = W`

`Delta E_k = (mv_f^2)/2 = (mv_i^2)/2`

`v_f, v_i ` represent the final and initial velocities

`Delta E_k = 0.62/2((-2.5)^2 - 0^2)`

`Delta E_k = 0.31*6.25 => Delta E_k ~~ 1.9J`

c) You may use the following distance formula, such that:

`x = v_i*t + (1/2)at^2`

Since you need to evaluate the velocity at x = 0, yields:

`0 = v_i*t + (1/2)at^2`

You need to evaluate the acceleration at x = 0.0 m. You need to notice that, on the graph, the force that corresponds to 0.0m is F = 0.2N, hence, you may evaluate the acceleration, such that:

`F = m*a => a = F/m => a = 0.2/0.62 => a = 0.32m/s^2`

Replacing 0.32 for a in equation `0 = v_i*t + (1/2)at^2`   yields:

`0 = v_i*t + (1/2)0.32*t^2 => -v_i = 0.16*t`

You may evaluate the time, using the following equation, such that:

`v = v_0 + a*t => t = (2.5 - 0)/0.32 => t = 7.8 s`

`v_i = -0.16*7.8 => v_i = -1.24 m/s`

Hence, evaluating the velocity at `x = 0m` yields `v_i = -1.24` m/s.