Question

A single conservative force acts on a 4.50-kg particle within a system due to its interaction with the rest of the system. The equation Fx = 2x + 4 describes the force, whereFx is in newtons and x is in meters. As the particle moves along the x axis from x = 0.90 m to x = 5.15 m, calculate the following.

(a) the work done by this force on the particle? (J)

(b) the change in the potential energy of the system? (J)

(c) the kinetic energy the particle has at x = 5.15 m if its speed is 3.00 m/s at x = 0.90 m? (J)

Answer 1

Answer: (a) The work done by this force on the particle is 42.71 J.

(b) The change in the potential energy of the system is -42.71 J.

(c) The kinetic energy the particle is 62.96 J.

Explanation:

(a)  For the given situation, expression for work done is as follows.

           W = $\int_{0.9}^{5.15}Fdx$

               = $\int_{0.9}^{5.15}(2x + 4)dx$

               = $[2\frac{x^{2}}{2} + 4x]^{5.15}_{0.9}$

               = $(x^{2} + 4x)^{5.15}_{0.9}$

               = $[(5.15)^{2} + 4(5.15) - (0.9)^{2} - 4(0.9)]$

               = 26.52 + 20.6 - 0.81 - 3.6

               = 42.71 J

Hence, the work done by this force on the particle is 42.71 J.

(b)  Expression for a conservative force is as follows.

              F = $-\frac{dU}{dx}$

            dU = -Fdx

      $\int_{0.9}^{5.15}dU$ =  $\int_{0.9}^{5.15}Fdx$

        $\int_{0.9}^{5.15}dU$ = -42.71 J

Therefore, the change in the potential energy of the system is -42.71 J.

(c) According to the work energy theorem,

         W = $\Delta K.E$

     $K.E_{0.9} - K.E_{5.15}$ = W

            $K.E_{0.9} = W + K.E_{5.15}$

                          = $42.71 + \frac{1}{2}mu^{2}$  

where, u = velocity of the mass at x = 0.9 m

         u = 3.0 m/s,       m = 4.50 kg

As,   $K.E_{0.9} = W + K.E_{5.15}$

                   = $42.71 + \frac{1}{2} \times 4.50 \times (3)^{2}$

                   = 62.96 J

Therefore, the kinetic energy the particle is 62.96 J.