Question

the stone of mass2kg is falling from the rest from the step hill what will becets kinetic energy after9sec

Answer 1

Since the slope of the hill is not given, we'll first assume it to make an angle of 45° with the ground & also the shape of stone to be cylindrical or spherical & is rolling frictionless from the hill.

The weight of the stone will act downwards, for our convenience, we'll resolve it in two perpendicular vectors, one parallel to the path of the stone & one perpendicular to it. The perpendicular vector will be cancelled out by the normal force exerted by the hinn on the stone, thus the motion of the stone will be because of the parallel vector, i.e. 2gcos45°=2√2g=√2g

Now, the velocity of the stone can be calculated by kinematical equation

v=u+at

Where, v is final velocity, u is initial velocity ( which here is 0m/s), a is acceleration (which here is √2gm/s2) & t is time (which here is 5s)

∴v=0+√2g(5)=5√2gm/s

If we put the value of g=9.8m/s2, we get

v=9.8×5×√2=69.29646455628m/s

Now we'll assume the slope of the hill to be perfectly vertical. Here,the hill will not exerted any force on the stone, the shape of the stone will not matter in this case.

Since the stone will fall vertically, it will have acceleration due to gravity unaffected. Thus, the velocity of stone will be:-

v=u+at=0+g(5)=5gm/s=49m/s

hope dis helpd u