Question

3/122 The collar has a mass of 2 kg and is attached to the light spring, which has a stiffness of 30 N/m and an unstretched length of 1.5 m. The collar is released from rest at A and slides up the smooth rod under the action of the constant 50‐N force. Calculate the velocity v of the collar as it passes position B.

Answer 1

Explanation:

According to the law of conservation of energy, work done by the force is as follows.

        $W_{F} = F Cos (30^{o}) \times 1.5$

                   = 64.95 J

Now, gain in potential energy is as follows.

                P.E = mgh

                      = $2 \times 9.8 \times 1.5$

                      = 29.4 J

Gain in potential energy will be as follows.

           = $\frac{1}{2}kx^{2}_{2} - \frac{1}{2}kx^{2}_{1}$

           = $\frac{1}{2} \times 30 N/m \times [(2.5 - 1.5)^{2} - (2 - 1.5)^{2}]$

           = 11.25

As,

          $W_{f} = u_{1} + u_{2} + \frac{1}{2}mv^{2}$

          $\frac{1}{2}mv^{2} = W_{f} - u_{1} - u_{2}$  

                   = 64.95 J - 29.4 - 11.25

                   = 24.3

              $v^{2} = \frac{24.3 \times 2}{2}$

                  v = 4.92 m/s

Therefore, we can conclude that relative velocity at point B is 4.92 m/s.