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 the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.

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;

= 1/2×k( - )

where k is spring constant ( 3 lb/in )

we substitute  

= 1/2 × 3( (-8)² - (5)² )      

= 1/2 × 3( 64 - 25 )

= 1/2 × 3( 39 )

= 58.5 lb-in

we convert to pound force-foot

= 58.5 × 0.0833333 lb-ft

= 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;

= -mgsin∅( x₂ - x₁ )        

we substitute in of values from the image below;

= -14 × sin(15°)( 5 - (-8) )  

= -14 × 0.2588 × 13

= -47.1  lb-in

we convert to pound force-foot

= -47.1 × 0.0833333 lb-ft

= - 3.935 lb-ft

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

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

the bal will not change its velocity.