First slide

Work

Question

A body with mass 2kg moves in one direction in the presence of a force which is described by the potential energy graph passing through (2,6) as shown. If the body is released from rest at x=2m, then it speed when it crosses x=5m is:

Moderate

A

zero
B

1m/s
C

2m/s
D

3m/s

Solution

$\begin{array}{l} From x = 2 m to x = 3.5 m \\ From, y = m x + c \\ s l o p e = \frac{0 - 6}{3.5 - 2} = - 4 \\ \frac{y_{2} - 0}{0 - 3.5} = - 4 \Rightarrow y_{2} = 14 \end{array}$
$\begin{array}{l} U_{1} = - 4 x + 14 \\ F_{1} = - \frac{d U_{1}}{d x} = + 4 N \\ W_{1} = + 4 \times (3.5 - 2) = + 6 J \\ From, x = 3.5 m to x = 4.5 m \\ \begin{array}{l} s l o p e = \frac{2 - 0}{4.5 - 3.5} = 2 \\ \frac{y_{2} - 0}{0 - 3.5} = 2 \Rightarrow y_{2} = - 7 \end{array} \\ U_{2} = + 2 x - 7 \\ F_{2} = - \frac{d U_{2}}{d x} = - 2 N \\ W_{2} = - 2 \times (4.5 - 3.5) \\ W_{2} = - 2 J \end{array}$
Work done from $4.5 m to 5 m=0$
$\begin{array}{l} \frac{1}{2} m v^{2} = W_{1} + W_{2} \\ \frac{1}{2} \times 2 \times v^{2} = 6 - 2 = 4 \\ \Rightarrow v = 2 m / s \end{array}$

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