A force vec{F_1} of magnitude 4 N acts on an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The components of force $\vec{F_1}$ are: $F_{1x} = 4 \cos 30^{\circ} = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3} \ N$ and $F_{1y} = 4 \sin 30^{\circ

Vedclass · Accepted Answer

$6.9$

Vedclass · Answer

(A) The components of force $\vec{F_1}$ are: $F_{1x} = 4 \cos 30^{\circ} = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3} \ N$ and $F_{1y} = 4 \sin 30^{\circ} = 4 	imes \frac{1}{2} = 2 \ N$.The components of force $\vec{F_2}$ are: $F_{2x} = 0 \ N$ and $F_{2y} = 4 \ N$.The net force components are: $F_x = F_{1x} + F_{2x} = 2\sqrt{3} \ N$ and $F_y = F_{1y} + F_{2y} = 2 + 4 = 6 \ N$.The magnitude of the resultant force is $F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{12 + 36} = \sqrt{48} \approx 6.928 \ N$.Using Newton's second law,the acceleration $a = \frac{F_R}{m} = \frac{6.928}{1} \approx 6.9 \ m \ s^{-2}$.

Similar Questions

$A$ body,under the action of a force $\vec F = 6\hat i - 8\hat j + 10\hat k$,acquires an acceleration of $1\, m/s^2$. The mass of this body must be

The force required to stop a body of mass $10 \,kg$ moving along a straight line path with a velocity of $10 \,ms^{-1}$ in a time of $10 \,s$ is (in $\,N$)

$A$ person holds a ball of mass $0.25 \ kg$ in his hand and throws it,so that it leaves his hand with a speed of $12 \ m \ s^{-1}$. In this process,if his hand moved through a distance of $0.9 \ m$,then the net force acted on the ball is (in $N$)

$A$ constant force acts on a body of mass $0.9 \,kg$ at rest for $10 \,s$. If the body moves a distance of $250 \,m$,the magnitude of the force is.........$N$

$A$ particle of mass $m$ is at rest at the origin at time $t = 0$. It is subjected to a force $F(t) = F_0e^{-bt}$ in the $x$ direction. Its speed $v(t)$ is depicted by which of the following curves?

Vedclass Test Series

Exam Paper Generator

Online Exam Module