Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)

Subsequently,

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,

According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4
.

Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
Answer:
U = -3978.8 J
Explanation:
The work of the gravitational force U just depends of the heigth and is calculated as:
U = -mgh
Where m is the mass, g is the gravitational acceleration and h the alture.
for calculate the alture we will use the following equation:
h = L-Lcos(θ)
Where L is the large of the rope and θ is the angle.
Replacing data:
h = 12.2-12.2cos(58.4)
h = 5.8 m
Finally U is equal to:
U = -70(9.8)(5.8)
U = -3,978.8 J
<u>Answer:
</u>
Cat has 2.02 seconds to right itself.
<u>Explanation:
</u>
Initial height of cat from ground = 20 meter.
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8
, we need to calculate time when s = 20 meter.
Substituting
So, cat has 2.02 seconds to right itself.
Answer: The motions shifts.