Suppose a cart is moving without any loss of energy to friction or air resistance. As the cart passes under a bridge, a heavy block is dropped vertically downward into the cart from a very small height above the cart. What happens to the speed of the cart

Initially the whole kinetic energy was with the mass of the cart now when we add some of the mass to the cart it's mass will increase but the energy will be same as that of initially of the lighter cart thus it's speed decrease to keep the kinetic energy constant.

Mathematically

[tex]K.E_{1}=\frac{1}{2}m_{cart}v^{2}\\\\K.E_{2}=\frac{1}{2}(m_{cart}+m_{block})v_{2}^{2}\\\\\therefore K.E_{1}=K.E_{2}\\\\\Leftrightarrow v_{2}=\sqrt{\frac{m_{cart}}{(m_{cart}+m_{block})}}v\\\\\therefore v_{2}< v[/tex]

fichoh fichoh · Answer 2 · 2021-11-12T07:30:24+01:00

Mass is inversely proportional to velocity, Hence, the velocity of the cart will decrease due to the additional mass of the block.

Comparing the kinetic energy of the cart before and after the block was dropped into the cart ;

Since there is no loss in energy ; then the kinetic energy is the same, since energy is conserved :

Recall :

Kinetic Energy = 0.5mv²

Let

mass of cart = 15 kg ; mass of block = 5 kg
K.E = 100 joules

Cart alone :

100 = 0.5(15)v²

100 = 7.5v²

v² = 100/7.5

v² = 13.333

v = √13.333 = 3.65 m/s

Cart and block :

100 = 0.5(15+5)v²

100 = 10v²

v² = 100/10

v² = 10

v = √10 = 3.16 m/s

Therefore, the speed of the cart will decrease due to the added mass of the block.

Learn more :https://brainly.com/question/11620508