A block of mass 5Ji kg is sitting on a frictionless ramp with a spring at the bo

Question

A block of mass 5Ji kg is sitting on a frictionless ramp with a spring at the bottom that has a spring constant of 570 N/m (refer to the figure). The angle of the ramp with respect to the horizontal is 26 degree. Part (a) The block, starting from rest, slides down the ramp a distance 69 cm before hitting the spring. How far, in centimeters, is the spring compressed as the block comes to momentary rest? 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? 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)

Let the distance it compressess the spring be x

So, all the gravitational PE stored at the top is converted to spring PE

So, mg(d+x)*sin(26) = 0.5*k*x^2

So, 5.8*9.8*(0.69 + x)*sin(26 deg) = 0.5*570*x^2

So, x = 0.293 m

So, change in gravitational PE = -0.5*k*x^2

= -0.5*579*0.293^2

= -24.9 J

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