A block of mass 5 kg is placed on a smooth inclined plane making an angle of 3 0 ∘ with the horizontal. It is connected by a light string passing over a smooth pulley at the top of the incline. The ot

Question

A block of mass 5 kg is placed on a smooth inclined plane making an angle of 3 0 ∘ with the horizontal. It is connected by a light string passing over a smooth pulley at the top of the incline. The other end of the string is attached to a spring of spring constant 500 N/m , which is fixed to a vertical wall. The system is in equilibrium. If the acceleration due to gravity is 10 m/s 2 , what is the extension (in cm) in the spring?

TestHub · Accepted Answer

For the block to be in equilibrium on the smooth inclined plane, the tension in the string must balance the component of gravity along the incline. $T = M g sin θ$ Substituting the given values: $M = 5 kg$ , $g = 10 m/s^{2}$ , $θ = 3 0^{\circ}$ . $T = 5 \times 10 \times sin (3 0^{\circ}) = 50 \times \frac{1}{2} = 25 N$

This tension is provided by the spring. According to Hooke's Law, $T = k x$ , where $k$ is the spring constant and $x$ is the extension. $x = \frac{T}{k} = \frac{25 N}{500 N/m} = 0.05 m$ Converting to centimeters: $x = 0.05 \times 100 = 5 cm$ .

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