A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is :
To determine the maximum spring compression upon impact with the block, we will apply the principle of energy conservation. This involves accounting for gravitational potential energy, the energy dissipated by friction, and the potential energy stored in the spring.
The block begins at rest, thus possessing zero initial kinetic energy. Its initial gravitational potential energy at the incline's apex is calculated as: \(U_g = mgh\). Given parameters are \(m = 5 \, \text{kg}\), \(g = 10 \, \text{m/s}^2\), and \(h = 10 \sin 30^\circ \, \text{m}\).
The height is computed as: \(h = 10 \times \frac{1}{2} = 5 \, \text{m}\).
Substituting these values into the potential energy equation yields: \(U_g = 5 \times 10 \times 5 = 250 \, \text{J}\).
As the block descends and traverses a frictional surface (\(\mu = 0.5\)) for \(2 \, \text{m}\), the work done by friction is: \(W_f = \mu mg d = 0.5 \times 5 \times 10 \times 2 = 50 \, \text{J}\).
According to energy conservation, the initial gravitational potential energy is transformed into spring potential energy and work done against friction: \(U_g = \frac{1}{2} k x^2 + W_f\).
Substituting the known values into the conservation equation: \(250 = \frac{1}{2} \times 100 \times x^2 + 50\).
Solving for \(x\):
The equation simplifies to: \(250 = 50 + 50 x^2\).
Rearranging to isolate \(x^2\): \(200 = 50 x^2\).
Dividing both sides by 50: \(x^2 = 4\).
The value of \(x\) is found to be: \(x = 2\).
Consequently, the spring experiences a maximum compression of \(2 \, \text{m}\) when the block impacts it.
