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It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibilty can be described by Hooke's law, which is depressed 0.65 cm for its maximum load of 115 kg. What is the spring's effective spring constant?

A player stands on the scale, and it depresses by 0.39cm. What is the mass of the player?,

Respuesta :

Kalahira
k = 1.7 x 10^5 kg/s^2 Player mass = 69 kg Hooke's law states F = kX where F = Force k = spring constant X = deflection So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration. F = kX F/X = k 115 kg* 9.8 m/s^2 / 0.65 cm = 115 kg* 9.8 m/s^2 / 0.0065 m = 1127 kg*m/s^2 / 0.0065 m = 173384.6154 kg/s^2 Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2 Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So X/0.39 cm = 115 kg/0.65 cm X = 44.85 kg/0.65 X = 69 kg The player masses 69 kg.
asifejaz20

Answer:

a) Spring constant = k = 1.73 × 10^5 N/m

b) Mass of player = m = 69 kg

Explanation:

a) According to hook’s law, "force applied on a spring is directly proportional to extension produced in the spring". So,  

                                        F = kx  

                       k = mg/x = (115×9.8)/0.0065 = 1.73 × 10^5 N/m

a) Mass of player = m = kx/g = (1.73 × 10^5)(0.0039)/9.8 = 69 kg.

Hence, player can play in under 85 kg team.

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