A body of mass m kg slides from rest along t | vedclass

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

Question

(B) Apply the Work-Energy Theorem $(WET)$ from point $A$ to point $B$: $W_{mg} = K_{B} - K_{A}$ Since the path is frictionless,the work done by gravit

Vedclass · Accepted Answer

$21.9$

Vedclass · Answer

(B) Apply the Work-Energy Theorem $(WET)$ from point $A$ to point $B$: $W_{mg} = K_{B} - K_{A}$ Since the path is frictionless,the work done by gravity is equal to the change in potential energy: $mg 	imes h = \frac{1}{2} mv_{B}^2 - 0$ From the geometry,the vertical height $h$ dropped by the body from $A$ to $B$ is $h = R \sin(45^{\circ}) + R = \frac{R}{\sqrt{2}} + R$. Substituting the values: $mg 	imes R \left( \frac{1}{\sqrt{2}} + 1 ight) = \frac{1}{2} mv_{B}^2$ $gR \left( \frac{1 + \sqrt{2}}{\sqrt{2}} ight) = \frac{1}{2} v_{B}^2$ $v_{B}^2 = 2gR \left( \frac{1 + 1.4}{1.4} ight) = 2 	imes 10 	imes 14 	imes \left( \frac{2.4}{1.4} ight)$ $v_{B}^2 = 20 	imes 10 	imes 2.4 = 480$ $v_{B} = \sqrt{480} \approx 21.9 \ m/s$. Thus,option $B$ is correct.

$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

Similar Questions

$A$ heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when:

$A$ body of mass $m$ attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ($g =$ gravitational acceleration).

$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

The minimum horizontal speed with which a body must be projected, so that it goes around a smooth vertical circular track of radius $4 \,m$ is $(g=9.8 \,ms^{-2})$. (in $\,ms^{-1}$)

An object is at the top of a smooth sphere which is kept fixed. As the object slides down after being given a negligible side push,what is the behavior of the magnitude of the acceleration of the object during its motion until it reaches the ground?

Vedclass Test Series

Exam Paper Generator

Online Exam Module