Work — Question 1

Question 1

Only the force component along displacement contributes

For a constant force, use $W=\vec F\cdot\Delta\vec r$. A cart moves 5 m horizontally right while a constant 20 N pull acts $60^\circ$ above the horizontal. The cart has no vertical displacement.

1. Find the work of the pull.

2. Find the separate contributions from horizontal and vertical force components.

3. If the same force acts during a 2 m horizontal leftward displacement, find its work.

Question 1 - Solution

Only the force component along displacement contributes

(a) Take the dot product.

The governing relation is $$\boxed{W=\vec F\cdot\Delta\vec r=Fd\cos\theta}$$

$$W=20(5)\cos60^\circ=50\ \mathrm J.$$ The full force magnitude times distance would overcount.

(b) Inspect both displacement components.

Resolve the force along the displacement axes: $$F_x=20\cos60^\circ=10\ \mathrm N,\qquad F_y=20\sin60^\circ=10\sqrt3\ \mathrm N.$$ $$W_x=F_x\Delta x=10(5)=50\ \mathrm J,\qquad W_y=F_y\Delta y=0.$$ A nonzero transverse force contributes no work when its corresponding displacement vanishes.

(c) Reverse displacement, not force.

Now $$\Delta x=-2\ \mathrm m$$ and the force remains unchanged, so $$W=10(-2)=-20\ \mathrm J.$$ Work is a signed scalar.