At the top:
Potential Energy = (mass) x (gravity) x (height)
= (30 kg) x (9.8 m/s²) x (3 meters)
= 882 joules
At the bottom:
Kinetic Energy = (1/2) x (mass) x (speed)²
= (1/2) x (30 kg) x (3 m/s)²
= (15 kg) x (9 m²/s²)
= 135 joules .
He had 882 joules of potential energy at the top,
but only 135 joules of kinetic energy at the bottom.
Friction stole (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :



v = 2 m/s
Hence, their speed after collision is 2 m/s.
Assume that the small-massed particle is
and the heavier mass particle is
.
Now, by momentum conservation and energy conservation:


Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

So now, we see that
and
. So therefore, the smaller mass recoils out.
Hope this helps you!
Bye!