Full points needs to show calculations. I\'ll submit twice for double points. A1

Question

Full points needs to show calculations. I'll submit twice for double points.

A16 through A18 .The diagram below illustrates a mass m = 10.0 kg which slides down an inclined plane. The coefficient of kinetic friction between the mass and the sloping surface is muk = 0.120. At the bottom of the incline, the mass slides over a frictionless horizontal plane before colliding with and compressing a spring of spring constant k = 5.00 times 104 Nm-1. The spring obeys Hooke's Law. By using the work-energy theorem, calculate the kinetic energy and hence the velocity of the mass just as it touches the spring. The compressed spring now exerts a force on the mass that accelerates it and gives it a velocity directed back towards the slope. Calculate how far up the slope the mass will travel before it comes to rest for the second time. Calculate the compression of the spring just as the mass comes to rest.

Explanation / Answer

Initial potential energy (total energy) = mgh = 10*9.8*5 =490 J
Energy lost due to friction. =(coff of friction) mg (cos theta)*d

= 0.12*10 *9.8 cos 45 * %u221A (5^2+ 5^2) = 58.8 J
Energy when it touches the spring = 490 - 58.8 = 431.2 J
Kinetic energy gained = 0.5mv^2 = 431.2 J
v^2 =2* 431.2/10 = 86.24
v = 9.3 m/s
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A17)

When mass comes to rest
0.5kx^2 = 431.2
x^2 = 2*431.2/ 5*10^4 =0.017248
x =0.13m
A18)
Before climbing up the incline it had energy of 431.2 J
It is lost while doing work against friction and moving through a distance s
431.2 =(coff of friction) mg (cos theta)*s + mg (sin theta)s

431.2 = 0.12*10 *9.8* (cos 45) *s + 10 *9.8*( sin 45) * s
s = 5.556 m
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