(900) Problem 4: A block of mass 3.8 kg is sitting on a frictionless ramp with a

Question

(900) Problem 4: A block of mass 3.8 kg is sitting on a frictionless ramp with a spring at the bottom that has a spring constant of 555 N/m (refer to the figure). The angle of the ramp with respect to the horizontal is 26 ©theexpertta.com 33% Part(a) The block, starting from rest, slides down the ramp a distance 39 cm before hitting the spring. How far in centimeters is the spring compressed as the block comes to momentary rest? Grade Summary 090 100% Potential sin() cosO | Submissions Attempts remaining: 10 (090 per attempt) detailed view tan() cotan asin acoso atan acotan sinh( coshO cotanh0 Degrees Radians 0 END NO EL CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback. () 3300 Part (b) After the block comes to rest, the spring pushes the block back up the ramp. How fast, in meters per second. is the block moving right after it comes off the spring? 3300 Part (c) what is the change of the gravitational potential energy, in Joules. between the original position of the block at the top of the ramp and the position of the block when the spring is fully compressed?

Explanation / Answer

a)

Potential energy gets converted to spring energy. If the spring compresses x then

mg(x + 0.39m)sin = 1/2kx^2

3.8kg x 9.8m/s^2 x (X + 0.39m) * sin26º = 1/2 x 555N/m x X^2

16.325X + 6.367 = 277.5X^2

277.5X^2 - 16.325X - 6.367 0 = 0

after solving quadratic equation

x = -0.12 m (not possible

x = 0.184 m = 18.4 cm

b)

Spring energy becomes PE and KE:

1/2kx^2 = mgxsin + 1/2mv^2

1/2 x 555N/m (0.184m)^2 = 3.8kg x 9.8m/s^2 * 0.184m * sin26º + 1/2 x 3.8kg x v^2

9.395 = 3J + 1.9 x v^2

v = 1.834 m/s

c)

mgh = 3.8kg x 9.8m/s^2 x -(0.184m + 0.39m) x sin26º

mgh = - 9.395J

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