A 15.0-kg block rests on a horizontal table and is attached to one end of a mass

Question

A 15.0-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 m/s in 0.360 s. In the process, the spring is stretched by 0.170 m. The block is then pulled at a constant speed of 5.00 m/s, during which time the spring is stretched by only 0.0500 m. (a) Find the spring constant of the spring. N/m (b) Find the coefficient of kinetic friction between the block and the table.

Explanation / Answer

(a) 1.00 × 103 N/m (b) 0.340

1) The pendulum clock's period of vibration is 2(L/g). Any change in g will therefore affect the period. The spring/mass clock's period is 2(m/k); g is not a factor here, so it will be much more accurate at altitude.

2) a = F/m (5/.5) = .05k/15 - µ*g
and conservation of momentum gives
10*m = F*t 10*15 = (.05k - µg)*.5

Solve these 2 eqs for k & µ............

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