Whan an elastic string of natural length 2m is fixed at one end and hangs vertically supporting a particle of mass 4Kg at the other end. It stretches to a length of 2.8m. A horizontal force of 28N is applied gradually to the mass until it is once again in equilibrium. Calculated the length and the inclination to the vertical of the string. If possible with a diagram.

Question

Asked on April 10, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-04-10T07:05:41.735Z

Step 1: Calculate the spring constant (k) of the elastic string. Given: Natural length L_0 = 2 m Stretched length L_1 = 2.8 m Mass m = 4 kg Acceleration due to gravity g = 9.8 m/s^2 First, calculate the initial extension (x_1): x_1 = L_1 - L_0 = 2.8 m - 2 m = 0.8 m Next, calculate the weight (W) of the particle: W = mg = 4 kg × 9.8 m/s^2 = 39.2 N According to Hooke's Law, the tension in the string (which is equal to the weight in this vertical setup) is T = kx. So, W = kx_1: k = (W)/(x_1) = 39.2 N0.8 m = 49 N/m Step 2: Analyze the forces when the horizontal force is applied. When the horizontal force F_H = 28 N is applied, the particle is in equilibrium under three forces: 1. The weight W = 39.2 N acting vertically downwards. 2. The horizontal force F_H = 28 N acting horizontally. 3. The tension T in the string, acting along the string at an angle to the vertical. Let x_2 be the new extension of the string and L_2 be the new stretched length. The tension in the string is T = kx_2 = 49x_2. We resolve the forces into horizontal and vertical components. For vertical equilibrium: T = W 49x_2 = 39.2 (*) For horizontal equilibrium: T = F_H 49x_2 = 28 (**) Step 3: Calculate the inclination to the vertical (). Divide equation (*) by equation (): (49x_2 )/(49x_2 ) = (28)/(39.2) = (28)/(39.2) = (280)/(392) = (5)/(7) = ((5)/(7)) ≈ 35.537^ The inclination to the vertical of the string is 35.5^. Step 4: Calculate the new length of the string (L_2). From = (5)/(7), we can construct a right-angled triangle with opposite side 5 and adjacent side 7. The hypotenuse is sqrt(5^2 + 7^2) = sqrt(25 + 49) = sqrt(74). So, = (7)/(sqrt(74)). Substitute this value into equation (*): 49x_2 ((7)/(sqrt(74))) = 39.2 x_2 = 39.2 sqrt(74)49 × 7 x_2 = 39.2 sqrt(74)343 x_2 ≈ (39.2 × 8.6023)/(343) ≈ 0.9831 m The new stretched length L_2 is the natural length plus the extension: L_2 = L_0 + x_2 L_2 = 2 m + 0.9831 m L_2 = 2.9831 m The length of the string is 2.983 m. Diagram: A diagram illustrating the forces and the inclination: ` Fixed point | | | Tension (T) | / |/ θ (angle to vertical) O----- F_H (Horizontal force) | | | W (Weight) | Particle `

Accepted Answer · 2026-04-10T07:05:41.735Z

Step 1: Calculate the spring constant (k) of the elastic string. Given: Natural length L_0 = 2 m Stretched length L_1 = 2.8 m Mass m = 4 kg Acceleration due to gravity g = 9.8 m/s^2 First, calculate the initial extension (x_1): x_1 = L_1 - L_0 = 2.8 m - 2 m = 0.8 m Next, calculate the weight (W) of the particle: W = mg = 4 kg × 9.8 m/s^2 = 39.2 N According to Hooke's Law, the tension in the string (which is equal to the weight in this vertical setup) is T = kx. So, W = kx_1: k = (W)/(x_1) = 39.2 N0.8 m = 49 N/m Step 2: Analyze the forces when the horizontal force is applied. When the horizontal force F_H = 28 N is applied, the particle is in equilibrium under three forces: 1. The weight W = 39.2 N acting vertically downwards. 2. The horizontal force F_H = 28 N acting horizontally. 3. The tension T in the string, acting along the string at an angle to the vertical. Let x_2 be the new extension of the string and L_2 be the new stretched length. The tension in the string is T = kx_2 = 49x_2. We resolve the forces into horizontal and vertical components. For vertical equilibrium: T = W 49x_2 = 39.2 (*) For horizontal equilibrium: T = F_H 49x_2 = 28 (**) Step 3: Calculate the inclination to the vertical (). Divide equation (*) by equation (): (49x_2 )/(49x_2 ) = (28)/(39.2) = (28)/(39.2) = (280)/(392) = (5)/(7) = ((5)/(7)) ≈ 35.537^ The inclination to the vertical of the string is 35.5^. Step 4: Calculate the new length of the string (L_2). From = (5)/(7), we can construct a right-angled triangle with opposite side 5 and adjacent side 7. The hypotenuse is sqrt(5^2 + 7^2) = sqrt(25 + 49) = sqrt(74). So, = (7)/(sqrt(74)). Substitute this value into equation (*): 49x_2 ((7)/(sqrt(74))) = 39.2 x_2 = 39.2 sqrt(74)49 × 7 x_2 = 39.2 sqrt(74)343 x_2 ≈ (39.2 × 8.6023)/(343) ≈ 0.9831 m The new stretched length L_2 is the natural length plus the extension: L_2 = L_0 + x_2 L_2 = 2 m + 0.9831 m L_2 = 2.9831 m The length of the string is 2.983 m. Diagram: A diagram illustrating the forces and the inclination: ` Fixed point | | | Tension (T) | / |/ θ (angle to vertical) O----- F_H (Horizontal force) | | | W (Weight) | Particle `

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