1. A spring and inclined plane are set up as shown below. The spring is compress

Question

1. A spring and inclined plane are set up as shown below. The spring is compressed 30.0 cm from its equilibrium position. At this point, it is 1.50 m from the bottom of the plane. The plane is at an angle of 30 degrees from horizontal. The mass is 0.500 kg. The coefficient of friction is 0.35. a) After the mass M is released, what is the speed of the mass when it reaches the bottom of the plane? b) What is the speed of the mass when it reaches the point 4.00 m up the inclined plane? (Hint: it?s not zero!) c) What is the total work done by friction?

Explanation / Answer

a)

Here,

let the speed of mass is u

Using work energy theorum

0.5mv^2 = 0.5kx^2 - uk*mg*d

0.5*0.5*v^2 = 0.5*500*0.30^2 - 0.35 * 0.5 * 9.8 * 1.5

solving for v

v = 8.93 m/s

the speed of block as the bottom of incline is 8.93 m/s

B)

at the top of incline ,

Using work energy theorum

0.5*0.5* (vf^2 - 8.93^2) = -.5 * 9.8*4*sin(30) - 0.35*.5*9.8*1.5*cos(30)

solving for vf

vf = 5.62 m/s

c)

total work done by friction = - 0.35 * 0.5 * 9.8 * 1.5 - 0.35*.5*9.8*1.5*cos(30)

total work done by friction = -4.8 J

the total work done by frcition is -4.8 J

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