The force exerted by the earth on a body is called ______of a body.
2 answers:
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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:
a) The work done by the applied force is 1500 joules.
b) The kinetic energy of the block after 10 seconds is 1200 joules.
c) The magnitude of the force of friction is 3 newtons and its direction is against motion.
d) 300 joules of energy are lost during motion.
Explanation:
a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (), measured in joules, is defined by the following expression:
(1)
Where:
- Force, measured in newtons.
- Distance, measured in meters.
If we know that and
, then the work done by the force exerted on the object is:
The work done by the applied force is 1500 joules.
b) At first we need to calculate the net acceleration of the object (), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:
(2)
Where is the initial velocity of the object, measured in meters per second.
We clear the acceleration within the equation above:
If we know that ,
and
, then the net acceleration experimented by the object is:
By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:
(3)
Where:
- External force exerted on the object, measured in newtons.
- Kinetic friction force, measured in newtons.
If we know that ,
and
, the kinetic friction force is:
The work done by friction (), measured in joules, is:
(4)
And the net work experimented by the object is:
By the Work-Energy Theorem we understand that change in translational kinetic energy (), measured in joules, is equal to the change in net work. That is:
(5)
If we know that , then the change in translational kinetic energy is:
The kinetic energy of the block after 10 seconds is 1200 joules.
c) The magnitude of the force of friction is 3 newtons and its direction is against motion.
d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.