Two masses of 1 , kg and 2 , kg respectively | vedclass

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

Question

(B) Let the acceleration of the $1 \, kg$ mass be $a_1 = 10 \, m/s^2$ (towards the right).Let the acceleration of the $2 \, kg$ mass be $a_2$.Applying

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let the acceleration of the $1 \, kg$ mass be $a_1 = 10 \, m/s^2$ (towards the right).Let the acceleration of the $2 \, kg$ mass be $a_2$.Applying Newton's second law for the system of two masses connected by a spring,the external force $F_{ext}$ acting on the system is equal to the sum of the products of mass and acceleration of each individual body:$F_{ext} = m_1 a_1 + m_2 a_2$Given $F_{ext} = 20 \, N$,$m_1 = 1 \, kg$,$m_2 = 2 \, kg$,and $a_1 = 10 \, m/s^2$:$20 = (1 	imes 10) + (2 	imes a_2)$$20 = 10 + 2 a_2$$10 = 2 a_2$$a_2 = 5 \, m/s^2$Thus,the acceleration of the $2 \, kg$ mass is $5 \, m/s^2$.

Similar Questions

The force constant of a spring is $K$. If half of the spring is cut off,what will be the force constant of the remaining spring?

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half length,the line $OA$ will

The figure shows two cases. In the first case,a spring (spring constant $K$) is pulled by two equal and opposite forces $F$ at both ends. In the second case,it is pulled by a force $F$ at one end while the other end is fixed. The extensions $(x)$ in the springs will be:

$A$ spring of force constant $k$ is cut into three equal pieces. If these three pieces are connected in parallel,the force constant of the combination will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module