Answer to Question #159431 in Mechanics | Relativity for Akinola Agboola

Question

Calculate the work done in displacing a body with a force,f=(30î + 40j)N through S=(2tî-4tj)m between t1=2 second and t2=5 seconds

Answer 1

By the definition of the work done, we have:

$$W=Fs$$

Let's first find displacements of the body at time $t_1=2\ s$ and $t_2=5\ s$:

$$d_1=(2\cdot2\hat{i}-4\cdot2\hat{j})=(4\hat{i}-8\hat{j})\ m,$$ $$d_2=(2\cdot5\hat{i}-4\cdot5\hat{j})=(10\hat{i}-20\hat{j})\ m.$$

The displacement of the body between $t_1=2\ s$ and $t_2=5\ s$ can be calculated as follows:

$$d=d_2-d_1=(10\hat{i}-20\hat{j})-(4\hat{i}-8\hat{j})=(6\hat{i}-12\hat{j})\ m.$$

Finally, we can find the work done between $t_1=2\ s$ and $t_2=5\ s$:

$$W=(30\hat{i}+40\hat{j})\ N\cdot(6\hat{i}-12\hat{j})\ m,$$ $$W=(30, 40)\cdot(6,-12)=30\cdot6+40\cdot(-12)=-300\ J.$$