Question

For Problems 26 and 27 The spring of constant $k = 50$ N/m is unstretched when the slider of mass $m = 2$ kg passes through position B. The slider is released from rest in position A. There is friction between slider and guide whose work done from A to B is -11 J and from B to C is -4 J. (Radius of guide R = 3 m). The whole system is in vertical plane.

Answer 1

70 N (This is example answer, the question is a setup for problems 26 and 27. The provided solution calculates the normal force at C.)

Answer 2

To solve this problem, we need to apply the Work-Energy Theorem. The total work done by non-conservative forces (friction in this case) equals the change in the total mechanical energy (kinetic + potential).

$W_{NC} = \Delta E_{mech} = (KE_f + PE_{grav,f} + PE_{spring,f}) - (KE_i + PE_{grav,i} + PE_{spring,i})$

Given data:

• Spring constant $k = 50$ N/m
• Mass of slider $m = 2$ kg
• Radius of guide $R = 3$ m
• Acceleration due to gravity $g = 10$ m/s$^2$ (standard assumption if not given for JEE/NEET)
• Slider released from rest at A, so $KE_A = 0$.
• Spring is unstretched at B, so $PE_{spring,B} = 0$.
• Work done by friction from A to B: $W_{f,AB} = -11$ J.
• Work done by friction from B to C: $W_{f,BC} = -4$ J.

1. Determine the spring's potential energy at A: The spring is unstretched at B. The slider is released from A, which is 4m above B. The diagram shows the spring attached to the slider and extending to the left at an angle. However, for the problem to be solvable with the given information, a common simplification is that the spring's extension/compression is directly related to the vertical displacement. This happens if the spring is vertical, or if its fixed end is very far away horizontally (making the angle small). Assuming the simplest interpretation that makes the problem solvable: the effective extension of the spring when the slider moves from B to A (or vice versa) is the vertical displacement of 4m. So, the extension of the spring at A is $\Delta x_A = 4$ m. The potential energy stored in the spring at A is: $PE_{spring,A} = \frac{1}{2} k (\Delta x_A)^2 = \frac{1}{2} (50 \text{ N/m}) (4 \text{ m})^2 = \frac{1}{2} (50)(16) = 25 \times 16 = 400$ J.

2. Calculate the speed of the slider at B ($v_B$): Let's set the reference level for gravitational potential energy at point B ($PE_{grav,B} = 0$). Then, the height of A above B is $h_{AB} = 4$ m. $PE_{grav,A} = mgh_{AB} = (2 \text{ kg})(10 \text{ m/s}^2)(4 \text{ m}) = 80$ J.

Applying the Work-Energy Theorem from A to B: $W_{f,AB} = (KE_B + PE_{grav,B} + PE_{spring,B}) - (KE_A + PE_{grav,A} + PE_{spring,A})$ $-11 \text{ J} = (\frac{1}{2} m v_B^2 + 0 + 0) - (0 + 80 \text{ J} + 400 \text{ J})$ $-11 = \frac{1}{2} m v_B^2 - 480$ $\frac{1}{2} m v_B^2 = 480 - 11 = 469$ J Since $m = 2$ kg: $\frac{1}{2} (2) v_B^2 = 469$ $v_B^2 = 469$ $v_B = \sqrt{469}$ m/s

3. Calculate the speed of the slider at C ($v_C$): Now, consider the motion from B to C. Let's set the reference level for gravitational potential energy at point C ($PE_{grav,C} = 0$). Point B is at a height $R$ above C. So, $h_{BC} = R = 3$ m. $PE_{grav,B} = mgR = (2 \text{ kg})(10 \text{ m/s}^2)(3 \text{ m}) = 60$ J. The spring is unstretched at B, so $PE_{spring,B} = 0$.

At C, the slider is at the lowest point of the circular guide. The spring's extension at C needs to be determined. From the diagram, if the fixed point of the spring is at the same horizontal level as B and at a distance $x_0$ to the left (so $L_0 = x_0$), then at C, the slider is vertically below B by a distance $R$. Slider at B: $(0, R)$ (relative to center of circle). Fixed point: $(-x_0, R)$. Length $L_0 = x_0$. Slider at C: $(0, -R)$. Length $L_C = \sqrt{x_0^2 + (R - (-R))^2} = \sqrt{x_0^2 + (2R)^2}$. Extension at C: $\Delta x_C = L_C - L_0 = \sqrt{x_0^2 + (2R)^2} - x_0$. This still depends on $x_0$.

If we continue with the interpretation that the spring is effectively vertical (meaning its extension is simply the vertical displacement relative to B's height), then: At B (height 0 for spring energy reference): $\Delta x_B = 0$. At C (height $-R$ relative to B): The spring would be compressed by $R$. So $\Delta x_C = R = 3$ m. $PE_{spring,C} = \frac{1}{2} k (\Delta x_C)^2 = \frac{1}{2} (50 \text{ N/m}) (3 \text{ m})^2 = \frac{1}{2} (50)(9) = 25 \times 9 = 225$ J.

Applying the Work-Energy Theorem from B to C: $W_{f,BC} = (KE_C + PE_{grav,C} + PE_{spring,C}) - (KE_B + PE_{grav,B} + PE_{spring,B})$ $-4 \text{ J} = (\frac{1}{2} m v_C^2 + 0 + 225 \text{ J}) - (469 \text{ J} + 60 \text{ J} + 0)$ $-4 = \frac{1}{2} m v_C^2 + 225 - 529$ $-4 = \frac{1}{2} m v_C^2 - 304$ $\frac{1}{2} m v_C^2 = 304 - 4 = 300$ J Since $m = 2$ kg: $\frac{1}{2} (2) v_C^2 = 300$ $v_C^2 = 300$ $v_C = \sqrt{300} = 10\sqrt{3}$ m/s.

4. Calculate the normal force at C ($N_C$): At point C, the slider is at the bottom of the circular path. The forces acting on it are:

• Gravitational force $mg$ downwards.
• Normal force $N_C$ upwards.
• Spring force $F_s$ upwards (since the spring is compressed). $F_s = k \Delta x_C = 50 \times 3 = 150$ N.

The net force towards the center of the circle (upwards) provides the centripetal force. $N_C + F_s - mg = \frac{m v_C^2}{R}$ $N_C + 150 - (2)(10) = \frac{(2)(300)}{3}$ $N_C + 150 - 20 = 200$ $N_C + 130 = 200$ $N_C = 200 - 130 = 70$ N.

The question itself is not provided, but these are the typical calculations required from such a problem statement. Assuming the question asks for the normal force at C.