A block of mass 5 kg is pulled by a force F as shown in the figure. If the coefficient of friction is 0.1, then the force needed to accelerate the block to 3 m / s 2 to the right is close to

Question

A block of mass 5 kg is pulled by a force F as shown in the figure. If the coefficient of friction is 0.1, then the force needed to accelerate the block to 3 m / s 2 to the right is close to (A) 12 N (B) 22 N (C) 32 N (D) 42 N

Tardigrade Admin · Accepted Answer

Component of force accelerating the block is

F cos θ

Net force on block,

F_{net} = F cos θ - μ N

= F cos 3 0^{\circ} - 0.1 \times 5 \times 9.8

= \frac{3 F}{2} - 4.9

From,

F_{net} = ma

, we have,

\frac{3}{2} F - 4.9 = 5 \times 3

\Rightarrow F = \frac{19.9}{3} = 22.9 N

Q. A block of mass 5kg is pulled by a force F as shown in the figure. If the coefficient of friction is 0.1, then the force needed to accelerate the block to 3m/s2 to the right is close to

12 N

22 N

32 N

42 N

Component of force accelerating the block is Fcosθ Net force on block, Fnet ​=Fcosθ−μN =Fcos30∘−0.1×5×9.8 =23​F​−4.9 From, Fnet ​=ma, we have, 23​​F−4.9=5×3 ⇒F=3​19.9​=22.9N

Q. A block of mass $5 k g$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is $0.1,$ then the force needed to accelerate the block to $3 m / s^{2}$ to the right is close to

Component of force accelerating the block is $F cos θ$
Net force on block, $F_{net} = F cos θ - μ N$
$= F cos 3 0^{\circ} - 0.1 \times 5 \times 9.8$
$= \frac{3 F}{2} - 4.9$
From, $F_{net} = ma$ , we have,
$\frac{3}{2} F - 4.9 = 5 \times 3$
$\Rightarrow F = \frac{19.9}{3} = 22.9 N$