# 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

###### Question:

## Answers

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

[tex]The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from[/tex]

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