Question

A block of mass 5 kg is placed on a rough horizontal plane. the coefficients of static energy kinetic friction between the block and surface as 0.5 and 0.3 respectively. A force of 100 N as shown in the figure. the acceleration of the block will be $g=10m/s^2$

• A. $(10-\sqrt{3})\sqrt{3}m/s^2$
• B. $20m/s^2$
• C. Zero
• D. $10\sqrt{3}m/s^2$

Answer 1

Answer: (A)

$(10-\sqrt{3})\sqrt{3}m/s^2$

Given force applied at an angle of ${30}^0$=100N

Resolving for horizontal component $F_x=F\cos \theta$

$F_x=100\frac{\sqrt{3}}{2}$

Now we know normal force $N=mg=5\times 10=50N$

and frictional force$f={\mu }_{s}N=0.3\times 50=15N$

Net force$F_{net}=F_x-f=\frac{100\sqrt{3}}{2}-15=\frac{100\sqrt{3}-30}{2}$

Again we know $F=ma$

$a=\frac{100\sqrt{3}-30}{2\times 5}=(10-\sqrt{3})\sqrt{3}m/s^2$