The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to

Missing figure and missing details can be found here:
http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....

Solution:
(a) The work done by the spring is given by
[tex]W= \frac{1}{2} k (\Delta x)^2 
[/tex]
where k is the elastic constant of the spring and [tex]\Delta x[/tex] is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
[tex]W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J[/tex]

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
[tex]W_W = -F_{//} (x_2 -x_1)[/tex]
where  the negative sign is given by the fact that [tex]F_{//}[/tex] points in the opposite direction of the displacement of the cart, and where
[tex]F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N[/tex]
therefore, the work done by the weight is
[tex]W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J[/tex]

This question is incomplete, the missing diagram is uploaded along this Answer below.

a) the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight is - 3.935 lb-ft

Explanation:

Given the data in the question;

(a) determine the work done on the cart by the spring

we calculate the work done on the cart by the spring as follows;

[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )

where k is spring constant ( 3 lb/in )

we substitute  

[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )      

[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )

[tex]W_{spring}[/tex] = 1/2 × 3( 39 )

[tex]W_{spring}[/tex] = 58.5 lb-in

we convert to pound force-foot

[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft

[tex]W_{spring}[/tex] = 4.875 lb-ft

Therefore, the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight

work done by its weight;

[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )        

we substitute in of values from the image below;

[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )  

[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13

[tex]W_{gravity}[/tex] = -47.1  lb-in

we convert to pound force-foot

[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft

[tex]W_{gravity}[/tex] = - 3.935 lb-ft

Therefore, the work done on the cart by its weight is - 3.935 lb-ft

does anyone know or what?

the journeys of marquette, joliet and la salle a) new france expand.

all three of these explorers explored the midwest and the area around the great lakes, which they subsequently claimed for france.

this was a big portion of territory and expanded new france which before then only consisted of quebec province in northeastern canada