(1) The potential energy at the top of the ball’s motion is 18 J.
(3) The kinetic energy increases as the potential energy decreases.
(4) The kinetic energy decreases as the potential energy increases.
(5) The total mechanical energy of the ball stays constant.
Explanation:
The total mechanical energy of the ball is equal to the sum of its kinetic energy (K, energy due to the motion) and its potential energy (U, energy due to the height of the ball). Mathematically:
In absence of friction, the mechanical energy of the ball is conserved, so in this case, it is always equal to 18 J. Let's now use this information to analyze each of the given statements:
(1) The potential energy at the top of the ball’s motion is 18 J. --> TRUE. In fact, at the top, the ball's speed becomes zero, so its kinetic energy is zero: K = 0. This means that all the mechanical energy of the ball is potential energy, therefore
E = U = 18 J
(2) The kinetic energy is less when the ball is thrown than when it is caught. --> FALSE. As we said, in absence of friction, the mechanical energy is conserved, therefore it always remains equal to 18 J.
(3) The kinetic energy increases as the potential energy decreases. --> TRUE. As we said, the sum of potential+kinetic energy remains constant:
E = K + U = 18 J
therefore, when the potential energy decreases, the kinetic energy increases.
(4)The kinetic energy decreases as the potential energy increases. --> TRUE. For the same reason described in (3).
(5)The total mechanical energy of the ball stays constant. --> TRUE. As we said at the beginning, the total mechanical energy is constant.
(6) The mechanical energy decreases as the ball moves up and increases as the ball comes down. --> FALSE. As we said, the mechanical energy remains constant, so it cannot change.
Learn more about kinetic and potential energy:
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Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Answer:
A) the maximum acceleration the boulder can have and still get out of the quarry
B) how long does it take to be lifted out at maximum acceleration if it started from rest
Explanation:
A)
let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.
the weight of the chain is: and maximum tension is
total mass and weight is :
∑
B)
maximum acceleration
using
to solve for t
Since static friction is the minimum force required to just start the motion of a stationary object.
Here if we need to start an object from rest then we required F = 700 N
So for the first part of the above problem Force will be F = 700 N
Now if the box is already moving then we will have to use kinetic friction force between box and floor
now we can write the equation of net force as
here
now we will have