(b) A 2.0 kg block is released from point A 0.96 m above the ground. The track i

Question

(b) A 2.0 kg block is released from point A 0.96 m above the ground.   

The track is frictionless up to point B; after B the rest of the track has a rough surface.

The block stops at point C; the distance between points B and C is 1.89 m.

Calculate

(i) the decrease in gravitational energy as the block goes from point A to point B.

(ii) the kinetic energy of the block at point B.

(iii) the speed of the block at point B

(iv) the work done by friction in bringing the block to rest.

(v) the friction force between B and C, fk.

(vi) the coefficient of kinetic friction between points B and C, mk.

.

Explanation / Answer

Ke at B = mgh = 2 x 9.8 x 0.96 = 18.8 J

Speed at B = sqrt 18.8 = 4.34 m/s

Work done by friction to bring it to rest W=Fs = 2x0.96x9.8 = 18.8 J

I'm assuming the block stops at C

So the work of friction equals the K at B

therefore F*x = 18.8J

so F(friction) = 18.8/1.89 = 9.96N

So ?*m*g = 9.96N or ? = 9.96/(2.0*9.8) = 0.508

Relevant formulas fk = uk n (although I don't have uk coefficient of kinetic friction)


*** But in later answer, I use uk = fk/ normal = 9.96 / 19.6(normal) = 0.51 uk ( this would be easy if i had this value to calculate fk) ***

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