Problem — Work of an inclined force

Work of a force. A force of $50\;\mathrm{N}$ is applied to a block over $5\;\mathrm{m}$. Calculate the work if the force is parallel to the displacement, then if it makes an angle of $60^\circ$ with it.

Solution

Data: $F = 50\;\mathrm{N}$, $s = 5\;\mathrm{m}$. The work is $W = F\,s\cos\theta$, where $\theta$ is the angle between the force and the displacement.

Case 1 — parallel force ($\theta = 0$):

$W = F\,s\cos 0^\circ = 50\cdot 5\cdot 1 = 250\;\mathrm{J}$

$\ev{W = 250\;\mathrm{J}}$

Case 2 — force at $60^\circ$:

$W = F\,s\cos 60^\circ = 50\cdot 5\cdot 0{,}5 = 125\;\mathrm{J}$

$\ev{W = 125\;\mathrm{J}}$

Physical reading. Only the component of the force along the displacement ($F\cos\theta$) does work; the perpendicular component contributes nothing. At $60^\circ$ the cosine is exactly $1/2$, so the work is halved compared with the parallel case, even though the force and displacement are identical in magnitude.

Collegamenti

Argomenti: Work and energy Concetti: Work of a force Competenze: Symbolic set-up Oggetti: Block