Question

Three forces act on block as shown in figure.

Find work done by net force, if the block moves $10\text{m}$ in horizontal direction
Given magnitudes of forces are
[$F_1=20\text{N},F_2=5\sqrt{2}\text{N},F_3=10\text{N}$]

• A. $50\text{J}$
• B. $100\text{J}$
• C. $75\text{J}$
• D. Work done is equal to zero as there is no friction.

Answer 1

The correct option is A $50\text{J}$

As net displacement is in horizontal direction, net force is also in horizontal direction.
$F_{\text{net}}=F_1\cos {60}^{\circ }+F_3\cos {90}^{\circ }-F_2\cos {45}^{\circ }$
$F_{\text{net}}=10\times \frac{1}{2}+0-5\sqrt{2}\times \frac{1}{\sqrt{2}}$
$F_{\text{net}}=5\text{N}$

Work done by net force $=F_{\text{net}}\times s$
$=5\times 10=50\text{J}$