Question 1:
moderateA force \(F = 20 + 10y\) acts on a particle in \(y\) direction where \(F\) is in newton and \(y\) in meter. Work done by this force to move the particle from \(y = 0\) to \(y = 1\text{ m})\) is:
(2019)
Work \(W = \int F dy\). Given \(F = 20 + 10y\). Limits \(y = 0\) to \(y = 1\text{ m})\). \(W = \int_0^1 (20 + 10y) dy = [20y + 5y^2]_0^1 = 20(1) + 5(1)^2 = 25\text{ J}\).
Question 2:
moderateA particle moves from a point \(P_1 = (-2\hat{i} + 5\hat{j})\) to \(P_2 = (4\hat{j} + 3\hat{k})\) when a force of \(F = (4\hat{i} + 3\hat{j})\text{ N})\) is applied. How much work has been done by the force?
(2016 – II)
Displacement \(dr = \vec{r_2} - \vec{r_1} = (2\hat{i} - \hat{j} + 3\hat{k})\). Force \(F = (4\hat{i} + 3\hat{j})\). Work \(W = F \cdot dr\). \(W = (4)(2) + (3)(-1) = 8 - 3 = 5\text{ J}\).
Question 3:
moderateA uniform force of \(F = (3\hat{i} + \hat{j})\text{ N})\) acts on a particle of mass \(2\text{ kg})\). Hence the particle is displaced from position \(\vec{r_1} = (2\hat{i} + \hat{k})\text{ m})\) to position \(\vec{r_2} = (4\hat{i} + 3\hat{j} – \hat{k})\text{ m})\). The work done by the force on the particle is:
(2013)
Displacement \(dr = \vec{r_2} - \vec{r_1} = (2\hat{i} + 3\hat{j} - 2\hat{k})\). Force \(F = (3\hat{i} + \hat{j})\). Work \(W = F \cdot dr\). \(W = (3)(2) + (1)(3) = 6 + 3 = 9\text{ J}\).
Question 4:
easyA body moves a distance of \(10\text{ m} \) along a straight line under the action of a \(5\text{ N}\) force. If the work done is \(25\text{ J}\), then the angle between the force and direction of motion of the body is
(1997)
Work \(W = Fd cos\theta\). Given \(W = 25\text{ J}\), \(F = 5\text{ N}\), \(d = 10\text{ m})\). \(25 = (5)(10) cos\theta\). \(cos\theta = \frac{1}{2}\). Therefore, \(\theta = 60^{\circ}\).
Question 5:
moderateA body, constrained to move in \(y\)-direction, is subjected to a force given by \(F = (-2\hat{i} + 15\hat{j} + 6\hat{k})\text{ N})\). The work done by this force in moving the body through a distance of \(10\hat{j}\text{ m})\) along \(y\)-axis, is:
(1994)
Force \(F = (-2hat{i} + 15hat{j} + 6hat{k})\). Displacement \(dr = 10hat{j}\text{ m})\). Work \(W = F cdot dr\). \(W = (15)(10) = 150\text{ J}\).
Question 6:
difficultA position dependent force, \(F = (7 – 2x + 3x^2)\text{ N})\) acts on a small body of mass \(2\text{ kg})\) and displaces it from \(x = 0\) to \(x = 5\text{ m})\). The work done in joule is:
(1994, 92)
\(W = int F dx\). Given \(F = 7 - 2x + 3x^2\). Limits \(x = 0\) to \(x = 5\text{ m})\). \(W = [7x - x^2 + x^3]_0^5 = 7(5) - 5^2 + 5^3 = 35 - 25 + 125 = 135\text{ J}\).
Question 7:
moderateThe K.E. of a person is just half of K.E. of a boy whose mass is just half of that person. If person increases its speed by \(1\text{ m/s}\), then its K.E. equals to that of boy then initial speed of person was:
(1999)
Let person's mass be \(M_p\) and speed \(v_p\). Boy's mass \(M_b = M_p/2\) and speed \(v_b\). Given \(KE_p = \frac{1}{2}KE_b\) and \(KE_p' = KE_b\) when \(v_p' = v_p+1\). From \(KE_p = \frac{1}{2}KE_b\), \(\frac{1}{2}M_p v_p^2 = \frac{1}{2} (\frac{1}{2} \frac{M_p}{2} v_b^2)\Rightarrow 4v_p^2 = v_b^2 \Rightarrow v_b = 2v_p\). From \(KE_p' = KE_b\), \(\frac{1}{2}M_p (v_p+1)^2 = \frac{1}{2}M_b v_b^2 = \frac{1}{2}(\frac{M_p}{2})(2v_p)^2 = \frac{1}{2}M_p (2v_p^2)\). So \((v_p+1)^2 = 2v_p^2 \Rightarrow v_p+1 = \sqrt{2}v_p\). \(1 = v_p(\sqrt{2}-1) \Rightarrow v_p = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1\text{ m/s}\).
Question 8:
easyTwo bodies of masses \(m\) and \(4m\) are moving with equal kinetic energies. The ratio of their linear momenta is:
(1998, 97, 89)
Kinetic energy \(KE = \frac{p^2}{2m}\), so momentum \(p = \sqrt{2mKE}\). Given \(KE_1 = KE_2 = KE\). For the two bodies, \(p_1 = \sqrt{2mKE}\) and \(p_2 = \sqrt{2(4m)KE}\). The ratio of their linear momenta is \(\frac{p_1}{p_2} = \frac{\sqrt{2mKE}}{\sqrt{8mKE}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\). So, the ratio is \(1:2\).
Question 9:
moderateThe potential energy between two atoms, in a molecule, is given by \(U(x) = \frac{a}{x^{12}} – \frac{b}{x^6}\) where \(a\) and \(b\) are positive constants and \(x\) is the distance between the atoms. The atom is in stable equilibrium, when:
(1995)
For equilibrium, the force is zero: \(F = -\frac{dU}{dx} = 0\). Given \(U(x) = ax^{-12} - bx^{-6}\), then \(\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7}\). Setting \(\frac{dU}{dx} = 0\) gives \(\frac{12a}{x^{13}} = \frac{6b}{x^7} \Rightarrow 12a = 6bx^6 \Rightarrow x^6 = \frac{12a}{6b} = \frac{2a}{b}\). So, \(x = (\frac{2a}{b})^{1/6}\). For stable equilibrium, \(\frac{d^2U}{dx^2} > 0\), which holds true for this value of \(x\).
Question 10:
moderateA particle is released from height \(S\) from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of Earth and the speed of the particle at that instant are respectively:
(2021)
Let the initial height be \(S\). At height \(h\), \(KE = 3PE = 3mgh\). By conservation of energy, \(mgS = mgh + 3mgh = 4mgh\), so \(h = S/4\). Also, \(KE = \frac{1}{2}mv^2 = 3mgh\), substituting \(h\) gives \(v^2 = 6g(S/4) = \frac{3gS}{2}\), so \(v = \sqrt{\frac{3gS}{2}}\).