A plank with a box on it at one end is gradua | vedclass

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

Question

(C) $1$. Coefficient of static friction $(\mu_s)$: When the box just starts to slip,the angle of inclination is the angle of repose. Thus,$\mu_s = 	a

Vedclass · Accepted Answer

$0.6$ and $0.5$

Vedclass · Answer

(C) $1$. Coefficient of static friction $(\mu_s)$: When the box just starts to slip,the angle of inclination is the angle of repose. Thus,$\mu_s = 	an 	heta = 	an 30^o = \frac{1}{\sqrt{3}} \approx 0.577 \approx 0.6$.$2$. Coefficient of kinetic friction $(\mu_k)$: The acceleration $a$ of the box sliding down the plank is given by $a = g \sin 	heta - \mu_k g \cos 	heta$.$3$. Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $S = 4.0\, m$,$u = 0$,and $t = 4.0\, s$:$4.0 = 0 + \frac{1}{2} a (4.0)^2 \Rightarrow 4.0 = 8a \Rightarrow a = 0.5\, m/s^2$.$4$. Substituting $a$ into the force equation (taking $g = 10\, m/s^2$):$0.5 = 10 \sin 30^o - \mu_k (10) \cos 30^o$$0.5 = 10(0.5) - \mu_k (10)(\frac{\sqrt{3}}{2})$$0.5 = 5 - 5\sqrt{3} \mu_k$$5\sqrt{3} \mu_k = 4.5$$\mu_k = \frac{4.5}{5 	imes 1.732} = \frac{0.9}{1.732} \approx 0.519 \approx 0.5$.Thus,$\mu_s \approx 0.6$ and $\mu_k \approx 0.5$.

Similar Questions

$A$ closed pipe is suddenly opened and changed to an open pipe of the same length. The fundamental frequency of the resulting open pipe is less than that of the $3^{rd}$ harmonic of the earlier closed pipe by $55 ~Hz$. Then,the value of the fundamental frequency of the closed pipe is: (in $~Hz$)

The compound$(s)$ with $P - H$ bond$(s)$ is$(are)$
$(A) \ H _3 PO _4$ $(B) \ H _3 PO _3$ $(C) \ H _4 P _2 O _7$ $(D) \ H _3 PO _2$

Which microorganism is used for the commercial production of citric acid?

In the circuit shown in the adjoining figure,the reading of the ammeter is .............. $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module