Unit 4: Work Energy and Power — Online MCQ Test
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This chapter covers: Work done by a constant force and a variable force kinetic and potential energies work-energy theorem power. The potential energy of a spring conservation of mechanical energy cons....
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Unit 4: Work Energy and Power — Important Questions & Answers
A bob of mass m is suspended at a point O by a light string of length l and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity at the point A, the string becomes slack when the bob reaches at the point D. The ratio of the kinetic energy of the bob at the points A and D is ______.
- A. 1
- B. 2
- C. 4
- D. 3
Answer: B. 2
At the point where the string becomes slack (D), the tension T = 0. mg*cos(theta) = mv_D^2/l. For the bob to reach D (assuming D is the highest point on the circular path relative to horizontal), v_D^2 = gl. Energy conservation between A and D: 1/2 m*v_A^2 = 1/2 m*v_D^2 + mg(h_D). Assuming D is at height R (relative to center) and A is bottom, h_D = l + l = 2l (if D is top) or l (if horizontal). Given standard JEE problem context where D is at level of suspension, height = l, v_D^2 = gl. KE_A = KE_D + mgl. If v_D^2=gl, KE_D = 1/2 mgl. KE_A = 1/2 mgl + mgl = 3/2 mgl. Ratio = 3.
At the point where the string becomes slack (D), the tension T = 0. mg*cos(theta) = mv_D^2/l. For the bob to reach D (assuming D is the highest point on the circular path relative to horizontal), v_D^2 = gl. Energy conservation between A and D: 1/2 m*v_A^2 = 1/2 m*v_D^2 + mg(h_D). Assuming D is at height R (relative to center) and A is bottom, h_D = l + l = 2l (if D is top) or l (if horizontal). Given standard JEE problem context where D is at level of suspension, height = l, v_D^2 = gl. KE_A = KE_D + mgl. If v_D^2=gl, KE_D = 1/2 mgl. KE_A = 1/2 mgl + mgl = 3/2 mgl. Ratio = 3.
A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by x(t) = alpha*t^2 + beta*t + gamma, where alpha = 1 m/s^2, beta = 1 m/s, and gamma = 1 m. The work done on the body during the time interval t = 2 s to t = 3 s is ____ J.
- A. 49
- B. 42
- C. 24
- D. 12
Answer: C. 24
x(t) = t^2 + t + 1. Velocity v(t) = 2t + 1. Acceleration a = 2 m/s^2. Force F = m*a = 2 * 2 = 4 N. Displacement S = x(3) - x(2) = (9+3+1) - (4+2+1) = 13 - 7 = 6 m. Work W = F*S = 4 * 6 = 24 J.
x(t) = t^2 + t + 1. Velocity v(t) = 2t + 1. Acceleration a = 2 m/s^2. Force F = m*a = 2 * 2 = 4 N. Displacement S = x(3) - x(2) = (9+3+1) - (4+2+1) = 13 - 7 = 6 m. Work W = F*S = 4 * 6 = 24 J.
Given below are two statements : Statement I: An object moves from position r1 to position r2 under a conservative force field F. The work done by the force is W = -∫[r1 to r2] F·dr. Statement II : Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force. In the light of the above statements, choose the correct answer from the options given below :
- A. Both Statement I and Statement II are true
- B. Statement I is false but Statement II is true
- C. Statement I is true but Statement II is false
- D. Both Statement I and Statement II are false
Answer: D. Both Statement I and Statement II are false
Statement I is incorrect because the work done by a conservative force is W = +∫[r1 to r2] F·dr (or negative change in potential energy). Statement II is incorrect because, by definition, the work done by a conservative force is independent of the path.
Statement I is incorrect because the work done by a conservative force is W = +∫[r1 to r2] F·dr (or negative change in potential energy). Statement II is incorrect because, by definition, the work done by a conservative force is independent of the path.
Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r = 0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is α m/s. Find α.
- A. 2
- B. 4
- C. 6
- D. 8
Answer: B. 4
By conservation of energy: Potential energy lost by 2m = Kinetic energy gained by m, 2m, and the rotational kinetic energy of the pulley. M_pulley = 30m. I = (1/2)M_pulley*r^2 = 15mr^2. mgh = (1/2)mv^2 + (1/2)(2m)v^2 + (1/2)I(v/r)^2. Solving for v yields 4 m/s.
By conservation of energy: Potential energy lost by 2m = Kinetic energy gained by m, 2m, and the rotational kinetic energy of the pulley. M_pulley = 30m. I = (1/2)M_pulley*r^2 = 15mr^2. mgh = (1/2)mv^2 + (1/2)(2m)v^2 + (1/2)I(v/r)^2. Solving for v yields 4 m/s.
Young's moduli of the material of wires A and B are in the ratio $1:4$, while its area of cross sections are in the ratio $1:3$. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires A and B will be in the ratio of [Assume length of wires A and B same]
- A. 1:12
- B. 1:36
- C. 12:1
- D. 36:1
Answer: D. 36:1
For a wire, elongation is $$\Delta L=\frac{FL}{AY}.$$ Here $F$ and $L$ are the same for both wires, so $$\Delta L \propto \frac{1}{AY}.$$ Given $Y_A:Y_B=1:4$ and $A_A:A_B=1:3$, we get $$\frac{\Delta L_A}{\Delta L_B}=\frac{A_BY_B}{A_AY_A}=\frac{3\times 4}{1\times 1}=12.$$ So the ratio is $12:1$? Wait, since elongation is inversely proportional to $AY$, wire A has smaller $AY$ than B, hence it elongates more. Therefore $$\frac{\Delta L_A}{\Delta L_B}=\frac{1/ (1\cdot 1)}{1/(3\cdot 4)}=12:1.$$ Thus the correct option is $12:1$.
For a wire, elongation is $$\Delta L=\frac{FL}{AY}.$$ Here $F$ and $L$ are the same for both wires, so $$\Delta L \propto \frac{1}{AY}.$$ Given $Y_A:Y_B=1:4$ and $A_A:A_B=1:3$, we get $$\frac{\Delta L_A}{\Delta L_B}=\frac{A_BY_B}{A_AY_A}=\frac{3\times 4}{1\times 1}=12.$$ So the ratio is $12:1$? Wait, since elongation is inversely proportional to $AY$, wire A has smaller $AY$ than B, hence it elongates more. Therefore $$\frac{\Delta L_A}{\Delta L_B}=\frac{1/ (1\cdot 1)}{1/(3\cdot 4)}=12:1.$$ Thus the correct option is $12:1$.