A block of mass 2 ,kg is connected to an idea | vedclass

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

Question

(D) Let $F$ be the tension in the spring. The potential energy $U$ of the spring is given by $U = \frac{1}{2} k x^2$, where $x$ is the extension of th

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(D) Let $F$ be the tension in the spring. The potential energy $U$ of the spring is given by $U = \frac{1}{2} k x^2$, where $x$ is the extension of the spring.The rate of change of potential energy is $\frac{dU}{dt} = kx \frac{dx}{dt} = F \cdot v_{rel}$, where $v_{rel}$ is the rate of change of extension, which is the relative velocity of the ends of the spring.Here, $v_{rel} = v_A - v_{block} = 6 \,m/s - 3 \,m/s = 3 \,m/s$.Given $\frac{dU}{dt} = 15 \,J/s$, we have $15 = F \cdot 3$, which gives $F = 5 \,N$.The force $F$ is the net force acting on the block of mass $m = 2 \,kg$.Using Newton's second law, $F = ma$, we get $5 = 2 \cdot a$.Therefore, $a = \frac{5}{2} = 2.5 \,m/s^2$.

Similar Questions

$A$ spring of force constant $k$ is cut into two equal halves. The force constant of each half is

Assertion $(A)$: When a spring is cut into two equal parts,the spring constant of each piece is twice that of the original spring. Reason $(R)$: Spring constant is inversely proportional to the length of the spring.

Two identical springs have the same force constant $73.5 \,Nm^{-1}$. The elongation produced in each spring in the three cases shown in Figure-$1$,Figure-$2$,and Figure-$3$ are (given $g=9.8 \,ms^{-2}$):

$A$ block of $3 \; kg$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in the figure. If spring $A$ is cut from the lower point at $t = 0$,find the acceleration of the block in $m/s^2$ at $t = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module