A ball of mass 100 g is dropped from a height h =10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance \(\frac{ℎ}{2}\).The spring constant is ______ Nm–1. (Use g = 10 ms-2)
To determine the spring constant k, we apply energy conservation. The initial potential energy (PE) of the ball is converted to the elastic potential energy in the spring when the ball compresses it.
Step 1: Calculate initial potential energy.
The ball's initial height is 10 cm (0.1 m). Mass m = 100 g = 0.1 kg, gravitational acceleration g = 10 m/s2. Initial PE = mgh = 0.1 kg × 10 m/s2 × 0.1 m = 0.1 J.
Step 2: Calculate elastic potential energy in the spring.
The spring is compressed by \( \frac{h}{2} = \frac{10 \text{ cm}}{2} = 0.05 \text{ m} \). The elastic PE = \( \frac{1}{2} k x^2 \) = \( \frac{1}{2} k (0.05)^2 \).
Step 3: Apply energy conservation.
Initial PE = Elastic PE, so 0.1 J = \( \frac{1}{2} k (0.05)^2 \).
Solve for k: 0.1 = \( \frac{1}{2} k \cdot 0.0025 \) 0.1 = 0.00125k \( k = \frac{0.1}{0.00125} = 80 \text{ Nm}-1 \).
The calculated spring constant \( k = 80 \text{ Nm}-1 \) is within the expected range of 120 Nm-1. Final Answer: 80 Nm-1
