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frozen [14]
3 years ago
14

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 he

ight 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
Physics
1 answer:
Likurg_2 [28]3 years ago
4 0

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.

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