Question 225: Work, Energy and Power & Circular Motion

Question

A 1 kg block subjected to two simultaneous forces $(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mathrm{N}$ and $(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \mathrm{N}$ is moved a distance of 25 m along $(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}})$ direction. The work done in this process is $\_\_\_\_$ J.

Accepted Answer

The two forces acting on the block are
$$ \vec F_1=(2\hat i+3\hat j+4\hat k)\,	ext{N} $$
and
$$ \vec F_2=(3\hat i-\hat j-2\hat k)\,	ext{N} $$
So, the resultant force is
$$ \vec F=\vec F_1+\vec F_2 $$
$$ \vec F=(2+3)\hat i+(3-1)\hat j+(4-2)\hat k $$
$$ \vec F=5\hat i+2\hat j+2\hat k $$
The block moves a distance of $25\,	ext{m}$ along the direction $(3\hat i-4\hat j)$.
First, find the unit vector in this direction.
Magnitude of $(3\hat i-4\hat j)$ is
$$ \sqrt{3^2+(-4)^2}=\sqrt{9+16}=5 $$
So the unit vector is
$$ \hat u=\frac{3\hat i-4\hat j}{5} $$
Hence, displacement vector is
$$ \vec s=25\hat u=25\left(\frac{3\hat i-4\hat j}{5}ight) $$
$$ \vec s=5(3\hat i-4\hat j) $$
$$ \vec s=15\hat i-20\hat j $$
Now, work done is
$$ W=\vec F\cdot \vec s $$
$$ W=(5\hat i+2\hat j+2\hat k)\cdot(15\hat i-20\hat j) $$
Using dot product,
$$ W=5(15)+2(-20)+2(0) $$
$$ W=75-40 $$
$$ W=35\,	ext{J} $$
So, the work done is
$$ \boxed{35\ 	ext{J}} $$

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type