The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car $A$ stops after 1000 m and car $B$ stops after 1500 m . If $F_A$ and $F_B$ are the forces applied by the breaks on cars $A$ and $B$ respectively, then the ratio of $\frac{F_A}{F_B}$ is
A bob of heavy mass $m$ is suspended by a light string of length $/$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:
An object moving along horizontal $$x$$-direction with kinetic energy $$10 \mathrm{~J}$$ is displaced through $$x=(3 \hat{i}) \mathrm{m}$$ by the force $$\vec{F}=(-2 \hat{i}+3 \hat{j}) \mathrm{N}$$. The kinetic energy of the object at the end of the displacement $$x$$ is
An object falls from a height of $$10 \mathrm{~m}$$ above the ground. After striking the ground it loses $$50 \%$$ of its kinetic energy. The height upto which the object can rebounce from the ground is:
At any instant of time $$t$$, the displacement of any particle is given by $$2 t-1$$ ($$\mathrm{SI}$$ unit) under the influence of force of $$5 \mathrm{~N}$$. The value of instantaneous power is (in $$\mathrm{SI}$$ unit):
A particle moves with a velocity $$(5 \hat{i}-3 \hat{j}+6 \hat{k}) ~\mathrm{ms}^{-1}$$ horizontally under the action of constant force $$(10 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+20 \hat{\mathrm{k}}) \mathrm{N}$$. The instantaneous power supplied to the particle is :
The potential energy of a long spring when stretched by $$2 \mathrm{~cm}$$ is U. If the spring is stretched by $$8 \mathrm{~cm}$$, potential energy stored in it will be :
The restoring force of a spring with a block attached to the free end of the spring is represented by
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms$$-$$1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 ms$$-$$2)
The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is

(Take g = 10 m s$$-$$2)


The work done on the particle during its displacement of 12 m is

on an object varies with distance x as shown here. The force is in N and x in m. The work done by the force in moving the object from x = 0 to x = 6 m is
