(c) Calculate the speed of the child at x = 0. Your response differs from the correct answer by more than 100%. m/s (d) Determine the value of x for which the kinetic energy of the system is a maximum. -0.01345 The response you submitted has the wrong sign. mm (e) Obtain the child's maximum upward speed. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

A 10 cm long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm. What is the spring constant k? How long is the spring when a 3.8 kg mass is suspended from it?

A bead of mass m = 5.30 kg is released from point A and slides on the frictionless track as shown in the figure below. The height of A is ha = 5.10 m. %3D ha 3.20 m 2.00 m (a) Determine the bead's speed at pbints B and C. point m/s В 8.4 point C Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four- digit accuracy to minimize roundoff error. m/s (b) Determine the net work done by the force of gravity in moving the bead from A to C. 141.12 Your response differs from the correct answer by more than 10%. Double check your calculations. J

Below is a conservation of energy problem. The solution to this problem is provided. Assess whether the solution provided is correct or incorrect AND EXPLAIN WHY. = 4.6 kg block is at rest against a horizontal spring that is compressed by 0.40 m. The spring has a spring constant of k₁ 2750 N/m. After leaving the spring, it travels up a 28° incline to a height of 2.8 m. At the top of the hill is a second spring with a spring constant of k₂ = 350 N/m. The horizontal portions are frictionless, but the hill has a coefficient of kinetic friction equal to uk = 0.16. The final velocity of the block is 9.78 m/s. How much is the second spring compressed by when the block comes to a stop against it? Simplifies to X₁ = 0.40 m h₁ = 0 m Vi = 0 m/s xf Wnc = 0 J 2 Wnc + mgh₁ +1/2 kx₁² +½ mv² = mgh₁+½ kxf² +½ mvf2 = k₁ = 2750 N/m k₂= 350 N/m m = 4.6 kg 2 ½ k₁x₁² = mghf + ½ K₂Xf2 2 ½ k₁x;² — mghƒ _½ k₂xf² Xf= 2750 N/m 350 N/m (0.4 m)2 x k 2 2gh -X f= X₁ = ? m h₁ = 2.8 m Vi = 0 m/s 0.141 m 2(9.8…