A spring, which has a spring constant of 1200 N/m, is compressed x = 0.20 m from

Question

A spring, which has a spring constant of 1200 N/m, is compressed x = 0.20 m from its equilibrium position with a 5.0 kg block and released at point A. The block travels to the right along a frictionless track and then up a 15° ramp to a horizontal platform at a height h = 0.30 m above the base.

a. What is the block’s velocity at point B?

b. The frictionless ramp is replaced with one that has a rough surface with a kinetic coefficient of friction of 0.30. The horizontal surfaces are still frictionless. How much work is done on the block by the frictional force as the block slides up the ramp to the platform?

c. With the rough ramp, what is the smallest value of the spring compression that will just get the block onto the platform?

Please give a thorough explanation for all work. Physics is a difficult subject for me and for me to succeed in this class I really need to understand how you arrived at each answer – thanks!

Explanation / Answer

we have, k= 1200N/m
x= 0.2 m
mass= 5 kg

a) total energy before = 1/2 kx2 = 1/2 x 1200 x 0.2 x 0.2 = 24 J

Now, after having a gain in potential energy of mgh = 5 x 9.8 x 0.3 J = 14.7 J

total kintic energy = 24 - 14.7 = 9.3 J = 1/2 x m x v2

thus, v = 1.92 m/s

b)Now, total length of slant surface = 1.12 m

and, in the slant length frictional force = mg cos 15 = 14.2 N

So, kinetic remaining at above slant surface = total kinetic energy before - loss due to frictional energy

But, we see here that it requires 14.2 x 1.12 + 14.7 = (friction loss in energy) + (gain in potential energy)

= 30.6 J of energy to get to plank above.

So, it will stop in the slant surface. It will not be able to go to the above horizontal surface.

c)we just calculated, we need 30.6 J to move the bolck just to the above plank.

So, 30.6 = 1/2 kx2

or, x = (121.2 /1200) = 0.32 meters.

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