The acceleration of the system of two bodies | vedclass

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

Question

(A) For a system of two masses $M$ connected by a string over a pulley on a wedge with angles $	heta_1 = 37^\circ$ and $	heta_2 = 53^\circ$,the acce

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) For a system of two masses $M$ connected by a string over a pulley on a wedge with angles $	heta_1 = 37^\circ$ and $	heta_2 = 53^\circ$,the acceleration $a$ is given by the net force divided by the total mass.The net driving force is $F_{net} = Mg \sin 53^\circ - Mg \sin 37^\circ$.The total mass of the system is $M + M = 2M$.Using Newton's second law,$F_{net} = (2M)a$.Therefore,$a = \frac{Mg \sin 53^\circ - Mg \sin 37^\circ}{2M} = \frac{g}{2} (\sin 53^\circ - \sin 37^\circ)$.Given $\sin 53^\circ \approx 0.8$ and $\sin 37^\circ \approx 0.6$,and $g = 10 \ m/s^2$:$a = \frac{10}{2} (0.8 - 0.6) = 5 	imes 0.2 = 1 \ m/s^2$.

The acceleration of the system of two bodies over the wedge as shown in the figure is ....... $m/s^2$. (Take $g = 10 \ m/s^2$)

Similar Questions

Two masses $M_1$ and $M_2$ are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination $\alpha$ and $\beta$. The tension in the string is:

$A$ truck of mass $10,000\, kg$ moves up an inclined plane rising $1$ in $50$ with a speed of $36\, km/h$. Find the power of the engine in $kW$ $(g = 10\, m/s^2)$.

$A$ solid sphere,a hollow sphere,and a ring are released from the top of an inclined plane (frictionless) so that they slide down the plane. Then,the maximum acceleration down the plane is for (no rolling):

The horizontal acceleration that should be given to a smooth inclined plane of angle $\sin^{-1} (1/l)$ to keep an object stationary on the plane relative to the inclined plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module