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S_A_V [24]
3 years ago
6

A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f

riction force of 15N. The block then slides up a frictionless ramp at 30 degrees reaching a 1.5m maximum height off the ground. What was the force constant of the spring
Physics
1 answer:
Setler [38]3 years ago
6 0

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

W_f = F_fs = 15*0.5 = 7.5 J

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

E_p = mgh = 0.1*10*1.5 = 1.5 J

Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

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Artyom0805 [142]

Answer:

option (c) is correct

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3 years ago
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Answer and Explanation:

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Answer:

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