An object of mass \(3text{ kg}\) is at rest. Now a force of \(vec{F} = 6that{i} + 4that{j}\) is applied on the object then velocity of object at \(t=3text{ second}\) is: (2002)
Given \(m = 3text{ kg}\), \(vec{F} = 6that{i} + 4that{j}\). Acceleration \(vec{a} = frac{vec{F}}{m} = frac{6t}{3}hat{i} + frac{4t}{3}hat{j} = 2that{i} + frac{4}{3}that{j}\). Velocity \(vec{v} = int vec{a} dt = int (2that{i} + frac{4}{3}that{j}) dt = t^2hat{i} + frac{2}{3}t^2hat{j}\), assuming initial velocity is zero. At \(t=3text{ s}\), \(vec{v} = (3^2)hat{i} + frac{2}{3}(3^2)hat{j} = 9hat{i} + 6hat{j}\). Note: There might be a slight discrepancy in the question's \(F_x\) component as presented versus options to obtain \(18hat{i}\). If \(F_x\) was \(12t\), then \(v_x\) would be \(18\).
A cricketer catches a ball of mass \(150text{ gm}\) in \(0.1text{ second}\) moving with speed \(20text{ ms}^{-1}\), then he experiences force of: (2001)
Force is the rate of change of momentum: \(F = frac{Delta p}{Delta t} = frac{m(v-u)}{Delta t}\). Given \(m = 150text{ gm} = 0.15text{ kg}\), \(u = 20text{ m/s}\), \(v = 0text{ m/s}\), \(Delta t = 0.1text{ s}\). So, \(F = frac{0.15 times (0 - 20)}{0.1} = frac{-3}{0.1} = -30text{ N}\). The magnitude of the force is \(30text{ N}\).
A force of \(6text{ N}\) acts on a body at rest and of mass \(1text{ kg}\). During this time, the body attains a velocity of \(30text{ m/s}\). The time for which the force acts on the body is: (1997)
Using Newton's second law, acceleration \(a = F/m = 6text{ N} / 1text{ kg} = 6text{ m/s}^2\). Since the body starts from rest (\(u=0\)) and attains a final velocity \(v=30text{ m/s}\), we use \(v = u + at\). So, \(30 = 0 + 6t\), which gives \(t = 30/6 = 5text{ s}\).
A \(10text{ N}\) force is applied on a body produce in it an acceleration of \(1text{ m/s}^2\). The mass of the body is: (1996)
From Newton's second law of motion, force \(F = ma\). Given \(F = 10text{ N}\) and \(a = 1text{ m/s}^2\). Therefore, mass \(m = F/a = 10text{ N} / 1text{ m/s}^2 = 10text{ kg}\).
A block has been placed on an inclined plane with the slope angle \(theta\), block slides down the plane at constant speed. The coefficient of kinetic friction is equal to: (1993)
When a block slides down an inclined plane at constant speed, the net force along the plane is zero. The component of gravitational force along the plane is \(mg sin \theta\). The kinetic friction force is \(mu_k mg cos \theta\). Equating these, we get \(mu_k = \tan \theta\).
Physical independence of force is a consequence of: (1991)
The principle of physical independence of forces states that each force acts independently without affecting other forces. This fundamental principle underpins the application of all Newton's laws of motion, especially when considering the net force as a vector sum of individual forces.
A particle of mass \(m\) is moving with a uniform velocity \(v_1\). It is given an impulse such that its velocity becomes \(v_2\). The impulse is equal to: (1990)
Impulse is defined as the change in momentum of an object. If the initial momentum is \(p_1 = mv_1\) and the final momentum is \(p_2 = mv_2\), then the impulse is \(I = p_2 - p_1 = m(v_2 - v_1)\).
A 600 kg rocket is set for a vertical firing. If the exhaust speed is \(1000 \text{ ms}^{-1}\,) the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is: (1990)
The thrust of a rocket is given by \(T = v_e \frac{dm}{dt}\,) where \(v_e\) is the exhaust speed and \(frac{dm}{dt}\) is the mass ejection rate. To overcome the rocket's weight \(Mg\), the thrust must equal the weight. So, \(v_e \frac{dm}{dt} = Mg\). Given \(M = 600 \text{ kg}\,) \(v_e = 1000 \text{ m/s}\,) and assuming \(g \approx 10 \text{ m/s}^2\), we have \(1000 \frac{dm}{dt} = 600 \times 10\). Thus, \(frac{dm}{dt} = 6 \text{ kg/s}\).
A block of mass \(m\) is placed on a smooth wedge of inclination \(theta\). The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (\(g\) is acceleration due to gravity) will be: (2004)
For no slipping, horizontal acceleration \(a = g tantheta\).The normal force \(N\) must balance the effective weight perpendicular to the incline.This gives \(N = sqrt{(mg)^2 + (ma)^2}\) if the forces are orthogonal.More correctly, resolving forces perpendicular to the incline, \(N - mg costheta - ma sintheta = 0\).Substituting \(a = g tantheta\), we get \(N = mg/costheta\).