hey!

so they have asked us about the total work done on the object

Work done=Force . distance travelled=F.s

and force can be calculated by F=m*a=3kg*20ms^-2=60N

now, work done=60N . 25m=1500Nm or J.

i think D should be the answer correct me if it’s not.

I also thought that d is the answer bit its C in the key

Hey!

Is the answer 1200?

We can divide this shape into two parts. The first part is from x=0 to x=15 and we’ll show it’s work with W1 . and the second part is from x=15 to x=25 and we’ll show it’s work with W2.

As we know the acceleration from x=0 until x=15 is constant. But from x=15 to x=25 we have variable acceleration.

W1=Fdcosx=madcosx=3×20×15×1=900

As we have variable acceleration in the second part we can’t consider the acceleration as a=20 , so we can find the average of the a which is 20+0/2=10.

So we can consider a=10 as the acceleration of the second part of the graph.

W2=Fdcosx=madcosx=3×10×10×1=300

W=W1+W2=900+300=1200