A basket and its contents have mass $M$. A monkey of mass $2M$ grabs the other end of the rope and very quickly (almost instantaneously) accelerates by pulling hard on the rope until he is moving with a constant speed of $v_{m/r} = 2ft/s$ measured relative to the rope. The monkey then continues climbing at this constant rate relative to the rope for $3$ seconds. How fast is the basket rising at the end of the $3$ seconds? Neglect the mass of the pulley and the rope. (given : $g = 32ft/s^2$)
$v_{basket} = 43\, ft/s$
$v_{basket} = 32\, ft/s$
$v_{basket} = 96\, ft/s$
$v_{basket} = 83\, ft/s$
Underline the correct alternative :
$(a)$ When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
$(b)$ Work done by a body against friction always results in a loss of its kinetic/potential energy.
$(c)$ The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
$(d)$ In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
A mass $m$ moving horizontally with velocity $v_0$ strikes a pendulum of mass $m$. If the two masses stick together after the collision, then the maximum height reached by the pendulum is
A body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{mv^2}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle
Assume the aerodynamic drag force on a car is proportional to its speed. If the power output from the engine is doubled, then the maximum speed of the car.
A 3.628 kg freight car moving along a horizontal rail road spur track at $7.2\; km/hour$ strikes a bumper whose coil springs experiences a maximum compression of $30 \;cm$ in stopping the car. The elastic potential energy of the springs at the instant when they are compressed $15\; cm$ is [2013]
(a) $12.1 \times 10^4\;J$ (b) $121 \times 10^4\;J$ (c) $1.21 \times 10^4\;J$ (d) $1.21 \times 10^4\;J$