Two masses of 10 , kg and 20 , kg respectivel | vedclass

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

Question

(B) Let $F_s$ be the spring force acting on the masses.For the $10 \, kg$ mass,the only horizontal force acting is the spring force $F_s$ towards the

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $F_s$ be the spring force acting on the masses.For the $10 \, kg$ mass,the only horizontal force acting is the spring force $F_s$ towards the right.Using Newton's second law,$F_s = m_1 a_1 = 10 \, kg 	imes 12 \, m/s^2 = 120 \, N$.Now,for the $20 \, kg$ mass,the forces acting are the applied force of $200 \, N$ towards the right and the spring force $F_s$ of $120 \, N$ towards the left.Applying Newton's second law for the $20 \, kg$ mass:$F_{net} = F_{applied} - F_s = m_2 a_2$$200 \, N - 120 \, N = 20 \, kg 	imes a_2$$80 \, N = 20 \, kg 	imes a_2$$a_2 = \frac{80}{20} = 4 \, m/s^2$.Therefore,the acceleration of the $20 \, kg$ mass is $4 \, m/s^2$ towards the right.

Similar Questions

$A$ uniform spring of force constant $k$ is cut into two pieces,the lengths of which are in the ratio $1 : 2$. The ratio of the force constants of the shorter and the longer pieces is

The system shown in the figure is in equilibrium and at rest. The spring and string are massless. Now,the string is cut. The acceleration of mass $2m$ and $m$ just after the string is cut will be:

$A$ spring with force constant $k$ is cut into two pieces. The length of one piece is twice the length of the other. What is the force constant of the larger piece?

The potential energy of a long spring when it is stretched by $3 \ cm$ is $U$. If the spring is stretched by $9 \ cm$,the potential energy stored in it will be: (in $U$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module