An ideal massless spring can be compressed 2.0 cm by a force of 270 N. A block w

Question

An ideal massless spring can be compressed 2.0 cm by a force of 270 N. A block whose mass is 12 kg is released from rest at the top of an incline as shown in Fig. 12-27, the angle of the incline being = 30°. The block comes to rest momentarily after it has compressed the spring by 6.2 cm.

An ideal massless spring can be compressed 2.0 cm by a force of 270 N. A block whose mass is 12 kg is released from rest at the top of an incline as shown in Fig. 12-27, the angle of the incline being 6- 30°. The block comes to rest momentarily after it has compressed the spring by 6.2 cm. 12 kg Figure 12-27 (a) How far has the block moved down the incline at this moment? (b) What is the speed of the block just as it touches the spring? m/s Additional Materials Section 12.1 Submit Answer Save Progress Practice Another Version

Explanation / Answer

The spring constant is:
k = 270 / 0.02 = 13500 N/m

When compressed by 6.2 cm, the stored potential energy is:
Ep = 13500 × 0.055^2 / 2 = 25.947 J

To have that much potential energy, the block must be at a height of:
25.947 = m×g×h = 12×9.8×H
H = 0.2206 m

The distance over the incline will thus be:
D = 0.2206/ sin(30) = 0.4412 m < - - - - - answer (a)

When just touching the spring, the vertical displacement of the block is
h = 0.2206 - 0.062×sin(30) = 0.1896 m

Thje potential energy for that height is:
Ep = 12×9.8×0.1896 = 22.2969 J

When converted to kinetic energy, the velocity will be:
22.2969 = 12×v²/2
v = 1.9277 m/s < - - - - - - - answer (b)

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