a) How far has the block moved down the incline at this moment? 0.485 Correct: Y

Question

a) How far has the block moved down the incline at this moment? 0.485 Correct: Your answer is correct. m

b) What is the speed of the block just as it touches the spring?

An ideal massless spring can be compressed 2.0 cm by a force of 270 N. A block whose mass is 12 kg is rele top of an incline as shown in Fig. 12-27, the angle of the incline being -30% The block comes to rest mo compressed the spring by 6.5 cm. 12 kg 30° Figure 12-27 (a) How far has the block moved down the incline at this moment? 0.485m (b) What is the speed of the block just as it touches the spring? oawhat s the speed of the block just as t touches the spring 2.180 X m/s

Explanation / Answer

k = 270 / 0.02 = 13500 N/m

When compressed by 6.5 cm, the stored potential energy is:
Ep = 13500 × 0.065² / 2 = 57.03 J

To have that much potential energy, the block must be at a height of:
57.03 = m×g×h = 12×9.8×H
H = 0.485 m (answer a)

The distance over the incline will thus be:
D = 0.485/ sin(30) = 0.97 m

When just touching the spring, the vertical displacement of the block is
h = 0.485 - 0.065×sin(30) = 0.4525 m

Thje potential energy for that height is:
Ep = 12×9.8×0.4525 = 53.21 J

When converted to kinetic energy, the velocity will be:
53.21 = 12×v²/2
v = 2.97 m/s ( answer (b))

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