RESPONDING TO AN ALARM, A 102 KG FIREMAN GETS UP FROM DINNER AND SLIDES DOWN A POLE TO THE GROUND FLOOR OF THE BUILDING, 3.3 M BELOW. THE FIREMEN STARTS AT REST AND LANDS ON THE GROUND FLOOR WITH A SPEED OF 4.2 M/S.
A.FIND THE FORCE EXERTED ON THE FIREMEN BY THE POLICE.
B.IF THE LANDING SPEED IS HALF THAT IN "PART A", IS THE FORCE EXERTED ON THE FIREMEN BY THE POLE DOUBLED? YES OR NO?
C.FIND THE FORCE EXERTED ON THE FIREMEN BY THE POLE WHEN THE LANDING SPEED IS 2.1 M/S.
To solve this problem, we can use the principles of work and energy.
A. FIND THE FORCE EXERTED ON THE FIREMAN BY THE POLE:
We can start by using the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The work done by the force of gravity is given by the equation:
Work = m * g * h
where m is the mass of the fireman, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the change in height.
In this case, the change in height is 3.3 m, so the work done by gravity is:
Work = 102 kg * 9.8 m/s^2 * 3.3 m = 3311.4 J
Since the fireman starts from rest, the initial kinetic energy is zero. The final kinetic energy is given by:
Final Kinetic Energy = (1/2) * m * v^2
where v is the final speed (4.2 m/s).
Plugging in the values, we get:
(1/2) * 102 kg * (4.2 m/s)^2 = 903.24 J
The work done by gravity is equal to the change in kinetic energy, so we can set them equal to each other:
3311.4 J = 903.24 J + FP * 3.3 m
where FP is the force exerted by the pole. Rearranging the equation, we can solve for FP:
FP = (3311.4 J - 903.24 J) / 3.3 m = 812.06 N
So, the force exerted on the fireman by the pole is approximately 812.06 N.
B. IF THE LANDING SPEED IS HALF THAT IN "PART A", IS THE FORCE EXERTED ON THE FIREMAN BY THE POLE DOUBLED? YES OR NO?
To answer this question, we need to consider the work-energy principle again. If the landing speed is halved, then the final kinetic energy will be one-fourth of the original value.
Let's denote the new force exerted by the pole as FPnew. According to the work-energy principle, we have:
FPnew * 3.3 m = (3311.4 J - 903.24 J/4)
Simplifying the equation, we get:
FPnew * 3.3 m = 2520.98 J
Solving for FPnew, we have:
FPnew = 2520.98 J / 3.3 m = 763.02 N
Comparing this value with the previous force (812.06 N), we can see that the force exerted on the fireman by the pole is not doubled when the landing speed is halved. Therefore, the answer is NO.
C. FIND THE FORCE EXERTED ON THE FIREMAN BY THE POLE WHEN THE LANDING SPEED IS 2.1 M/S.
To find the force exerted on the fireman by the pole when the landing speed is 2.1 m/s, we can use the same approach as before.
First, we need to calculate the final kinetic energy using the given landing speed:
Final Kinetic Energy = (1/2) * m * v^2
= (1/2) * 102 kg * (2.1 m/s)^2
= 231.41 J
We can then set the work done by gravity equal to the change in kinetic energy:
3311.4 J = 231.41 J + FP * 3.3 m
Rearranging the equation, we can solve for FP:
FP = (3311.4 J - 231.41 J) / 3.3 m
= 974 N
Therefore, the force exerted on the fireman by the pole when the landing speed is 2.1 m/s is approximately 974 N.