Question

A spring with a spring-constant 2.2 N/cm is compressed 28 cm and released. The 5 kg mass skids down the frictional incline of height 31 cm and inclined at a 18 angle. The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.8 m along the incline which has a coefficient of friction of 0.3. k = 2.2 N/cm What is the final velocity vf of the mass? Answer in units of m/s

Answer 1

Explanation:

Formula for work done by sphere is as follows.

        $W_{s} = \frac{1}{2} kx^{2}$

                   = $\frac{1}{2} 220 N/m \times (0.31 m)^{2}$

                    = 10.57 J

Formula for work by friction is as follows.

          $w_{f} = -fS$

                    = $-(\mu mg Cos \theta) S$

                    = $-(0.8 \times 5 kg \times 9.8 m/s^{2} \times Cos (18^{o})) \times 0.8 m$

                    = $-(0.8 \times 5 kg \times 9.8 m/s^{2} \times 0.951 \times 0.8 m$

                   = -29.82 J

Work by gravity is given as follows.

         $w_{g}$ = mgh

                     = $5 kg \times 9.8 m/s^{2} \times 0.31 m$

                     = 15.19 J

Hence, net work done will be given as follows.

          $w_{s} + w_{f} + w_{g}$

        = 10.57 J + (-29.82) J + 15.19 J

        = -4.06 J

According to work energy theorem,

               w = $\Delta k = \frac{1}{2}mv^{2}_{f}$

         -4.06 J = $\frac{1}{2} \times 5 kg \times v^{2}_{f}$

              $v_{f}$ = 1.27 m/s

Thus, we can conclude that final velocity of the given mass is 1.27 m/s.