High School

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A 10.1 kg uniform board is wedged into a corner and held by a spring. The spring has a spring constant of 176 N/m and is parallel to the floor. Find the amount by which the spring is stretched.

Answer :

To determine the stretch in the spring from a 10.1 kg board, you find the force due to the board's weight and equate it to the restoring force of the spring. Using Hooke's law and substituting the given values, we calculate that the spring is stretched by 56.2 cm.

The question has asked us to calculate the amount by which a spring is stretched due to the force exerted by a 10.1 kg uniform board that is wedged in a corner and held horizontally by the spring.

To find the stretch in the spring, we need to first determine the force exerted by the weight of the board, which is its mass multiplied by the acceleration due to gravity (9.8 m/s2). This force equals the restoring force provided by the spring, which can be described by Hooke's law (F = -kx, with k being the spring constant).

F = -kx

mg = kx

x = mg/k

Substituting the given values:

m = 10.1 kg (mass of the board)

g = 9.8 m/s2 (acceleration due to gravity)

k = 176 N/m (spring constant)

Now, we calculate:

x = (10.1 kg * 9.8 m/s2) / 176 N/m

x = 0.562 meters or 56.2 cm

Therefore, the spring is stretched by 56.2 cm to hold the board in equilibrium.

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