Q A block with mass m = 16 kg rests on a frictionless table and is accelerated b

Question

Q A block with mass m = 16 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 5247 N/m after being compressed a distance x1 = 0.408 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2 m long. For this rough path, the coefficient of friction is k = 0.41.

1) How much work is done by the spring as it accelerates the block?

J

2) What is the speed of the block right after it leaves the spring?

m/s

3) How much work is done by friction as the block crosses the rough spot?

J

4) What is the speed of the block after it passes the rough spot?

m/s

Explanation / Answer

1) Work done by the spring is W = 0.5*k*x^2 = 0.5*5247*0.408^2 = 436.7 J

2) Using Work energy theorem

Work done by the spring force = change in Kinetic energy

436.7 = 0.5*m*V^2

436.7 = 0.5*16*V^2

V = 7.38 m/sec

3) Work done by friction is Wf = f*S = mu_k*m*g*s = 0.41*16*9.8*2 = 128.576 J

4) Work done by the net force = change in kinetic energy

-128.576 =0.5*m*V^2 - 0.5*m*vo^2

-128.576 = 0.5*16*V^2 - (0.5*16*7.38^2)

V = 6.19 m/sec

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