Please make work legible and clear I want to understand how to do it. Please als

Question

Please make work legible and clear I want to understand how to do it. Please also only use Newtonian mechanics
6. A spring that obeys Hooke's Law, with spring constant k = 400 N/m lays along an incline, at 30° to the horizonta. A block of mass 5.0 kg is released 5.0 meters up to slope from the end of the spring, so that it will slide down onto the spring. The coefficient of kinetic friction between the block and the surface is = 0.1 (assume that the static friction coefficient is low enough that the block begins to slide immediately when released). (a) What is the kinetic energy of the block just before it collides with the spring? (b) How far is the spring compressed by the block? This distance is , in the dia- gram.) (c) After bouncing off the spring the block travels back up the slope. How far back up the slope does the block travel after bouncing off of the spring? (This distance is ae in the diagram.) 5.0 kg 5.0 m spring Xc 30

Explanation / Answer

a) When the block is released, it moves along the slope, its potential energy decreases and kinetic energy increases (because speed increases). One could say potential energy get converted into kinetic energy, but energy is lost due to friction too. We have to find KE just before it collides with spring.

Potential Energy decreased = Kinetic energy increased + energy lost due to friction

mg* (change in height ) = KEfinal + Ffric*displacement along frictional force

5kg*9.8m/s2 * (5m sin 30degree) = KEfinal + [0.1*5kg*9.8m/s2 *cos(30degree)]*5m

49 *5 *0.5 J = KEfinal + 4.9*0.866*5J

KEfinal = 122.5 J - 21.2J =101.3J

b) Now this kinetic energy will decrease as the block's speed decreases when it compresses spring. The KE is zero at maximum compression as speed becomes zero. This will be stored in spring potential energy. Meanwhile some gravitation potential energy will also decrease as block's verical height is decreasing

Initial KE + change in gravitational PE = increase in spring PE + energy lost due to Friction

101.3 J + 5kg*9.8m/s2 * (xb sin 30degree) = 1/2 k xb2 + [0.1*5kg*9.8m/s2 *cos(30degree)]*xb

101.3J + 24.5 xb = 0.5 *400* xb2 + 4.2*xb

200 xb2 -20.3xb -101.3 =0

Solving the quadratic equation we will get xb=0.76m, we will ignore negative value

c) Putting this value of xb we will find spring potential energy after compression = 116.7 J

Now when the block bounces, this spring energy get used up to gain some kinetic energy and some gravitational potential energy, but at highest point KE becomes zero again. So we will consider only gravitational PE and energy lost due to friction. Here x is the distance from the last compression point. Don't get it confused with Xc

116.7 =  5kg*9.8m/s2 * (x sin 30degree) + [0.1*5kg*9.8m/s2 *cos(30degree)]*xb

116.7=24.5x + 4.2x=28.7x

x=116.7/28.7=4.06m

xc=x-xb= 4.06-0.76m=3.3 m

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