A body with mass 2 kg moves in one direction in the presence of a force which is described by the potential energy graph. If the body is released from rest at x = 2 m, then its speed when it crosses x = 5 m is

Question

A body with mass 2 kg moves in one direction in the presence of a force which is described by the potential energy graph. If the body is released from rest at x = 2 m, then its speed when it crosses x = 5 m is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fd50fe7b6026341150217ee63030f369-.png" /> (A) zero (B) 1 ms-1 (C) 2 ms-1 (D) 3 ms-1

Tardigrade Admin · Accepted Answer

Let at x=2 m , U=U2 (10-U2/2-1)=(10-0/3.5-1) ⇒ U2=6 J (K E+P E)x=2 m =(K E+P E)x=5 m ⇒ 0+6=(1/2)(2) v2+2 ⇒ v=2 m / s

Q. A body with mass $2 k g$ moves in one direction in the presence of a force which is described by the potential energy graph. If the body is released from rest at $x = 2 m,$ then its speed when it crosses $x = 5 m$ is

zero

$1 m s^{- 1}$

$2 m s^{- 1}$

$3 m s^{- 1}$

Let at $x = 2 m, U = U_{2}$
$\frac{10 - U _{2}}{2 - 1} = \frac{10 - 0}{3.5 - 1}$
$\Rightarrow U_{2} = 6 J$
$(K E + PE)_{x = 2 m} = (K E + PE)_{x = 5 m}$
$\Rightarrow 0 + 6 = \frac{1}{2} (2) v^{2} + 2$
$\Rightarrow v = 2 m / s$

Q. A body with mass 2kg moves in one direction in the presence of a force which is described by the potential energy graph. If the body is released from rest at x=2m, then its speed when it crosses x=5m is

zero

1ms−1

2ms−1

3ms−1

Let at x=2m,U=U2​ 2−110−U2​​=3.5−110−0​ ⇒U2​=6J (KE+PE)x=2m​=(KE+PE)x=5m​ ⇒0+6=21​(2)v2+2 ⇒v=2m/s

Q. A body with mass $2 k g$ moves in one direction in the presence of a force which is described by the potential energy graph. If the body is released from rest at $x = 2 m,$ then its speed when it crosses $x = 5 m$ is

$1 m s^{- 1}$

$2 m s^{- 1}$

$3 m s^{- 1}$

Let at $x = 2 m, U = U_{2}$
$\frac{10 - U _{2}}{2 - 1} = \frac{10 - 0}{3.5 - 1}$
$\Rightarrow U_{2} = 6 J$
$(K E + PE)_{x = 2 m} = (K E + PE)_{x = 5 m}$
$\Rightarrow 0 + 6 = \frac{1}{2} (2) v^{2} + 2$
$\Rightarrow v = 2 m / s$