The K.E. at the maximum height is zero, no matter what that height is.
Answer:2.103 m/s
Explanation:
Given
mass of sports car 
mass of SUV 
Suppose u is the velocity if sports car before collision
Conserving momentum we get
After collision the combined mass drag 2.8 m and finally stops
From work energy theorem work done by friction is equal to change in kinetic energy of the combined mass system
where 
Initial velocity 
Answer:
m = 369 grams
Explanation:
Given that,
The density of backpack, d = 30 g/mL
The volume of the backpack, V = 12.3 cm³
We need to find the mass of the backpack. The density of an object is given by :
So, the mass of the backpack is 369 grams.
Answer:
Transfer of charge by touching is called conduction.
Wow ! This is not simple, Shoot, and I give you a lot of credit
and an extra merit badge if you're generally keeping up with it.
I scratched my head for a few minutes, and I think I've got it.
Here's what I think is going on:
KE₁ = KE of the box before pushing
(1/2) (m) (speed²) = 10 x 2² = 40 joules
KE₂ = KE of the box after pushing 3m
(1/2) (m) (speed²) = 10 x 4² = 160 joules
The box gained (160 - 40) = 120 J of kinetic energy.
Now look at the cluttered force diagram.
Cat's component of force in the direction of motion is 120N.
That's the part of her force that does the work on the box.
How much work does she do ?
(force) x (distance) = (120N) x (3m) = 360 joules .
Only 120 J of that energy showed up as increased kinetic energy
of the box. The other 240J of her hard-earned work was consumed
by friction.
Work of friction = (Friction force) x (distance)
240 J = (friction force) x (3 m)
240 J / 3 m = friction 'force' = 80 N .
I think that's it.
What I did was:
-- Find the work that Cat did.
-- Find the increase in the kinetic energy of the box.
-- The difference ... the 'missing energy' ... was the work done
by friction in the same distance.
Does this do anything for you ?
