Potential energy of a body of mass 1 kg free | vedclass

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

Question

(A) The total mechanical energy $E$ of a system is the sum of its potential energy $U$ and kinetic energy $KE$. That is,$E = U + KE$.Given,$E = 2 \ J$

Vedclass · Accepted Answer

$\sqrt{5} \ ms^{-1}$

Vedclass · Answer

(A) The total mechanical energy $E$ of a system is the sum of its potential energy $U$ and kinetic energy $KE$. That is,$E = U + KE$.Given,$E = 2 \ J$ and $m = 1 \ kg$.To find the maximum speed,we need the maximum kinetic energy,which occurs when the potential energy is at its minimum.$U(x) = \frac{x^2}{2} - x$.To find the minimum potential energy,we set $\frac{dU}{dx} = 0$:$\frac{d}{dx}(\frac{x^2}{2} - x) = x - 1 = 0 \implies x = 1 \ m$.$U_{\min} = U(1) = \frac{1^2}{2} - 1 = -0.5 \ J$.Since $E = U + KE$,we have $KE_{\max} = E - U_{\min} = 2 - (-0.5) = 2.5 \ J$.Using $KE = \frac{1}{2}mv^2$:$2.5 = \frac{1}{2}(1)v^2 \implies v^2 = 5 \implies v = \sqrt{5} \ ms^{-1}$.

Potential energy of a body of mass $1 \ kg$ free to move along $X$-axis is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the body is $2 \ J$,then the maximum speed of the body is (Assume only conservative force acts on the body).

Similar Questions

When an object is released,its potential energy $U$ decreases,and its velocity becomes $v$. What is the mass of the object?

$A$ uniform flexible chain of mass $m$ and length $2l$ hangs in equilibrium over a smooth horizontal pin of negligible diameter. One end of the chain is given a small vertical displacement so that the chain slips over the pin. The speed of the chain when it leaves the pin is

In the presence of which of the following is the work-energy theorem valid?

$A$ particle of mass $m$ moves with a velocity $v = a\sqrt{x}$,where $a$ is a constant and $x$ is the displacement. Find the total work done by all forces during the displacement from $x = 0$ to $x = d$.

$A$ body of mass $1 \text{ kg}$ moves along a straight line with a velocity $v = 2x^2$. The work done by the body during displacement from $x = 0$ to $5 \text{ m}$ is . . . . . . $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module