Question

The 8kg ball is connected with a spring. Initially, the spring is compressed by 0.2m. Then under the applied force F, the spring is stretched to 0.6m. Determine the total workdone to the ball during this process. The slope is smooth.

Answer 1

Given:

mass=8kg

spring is compressed= 0.2m

spring is stretched = 0.6m

Find the total work done to the ball during this process.

Now,

$\\Total \,elevation\, change\, of \,the\, ball\,(h)=0.6+0.2=0.8m\\[10pt] Work\,done\,to\,the\,ball\,=Change\,in \,potential\,energy\,of\,the\,ball\\[10pt] potential\,energy\,of\,the\,ball=(mgh)=(8kg\times 9.81m/s{^{2}}\times 0.8m)=62.78J\\[10pt] Hence\,, Work\,done\,to\,the\,ball\,=62.78J$