4. the fire alarm goes off, and a 97 kg fireman slides 3.0 m down a pole to the ground floor. suppose the fireman starts from rest, slides with a constant acceleration, and reaches the ground floor in 1.2 s. (a) what was the upward force f exerted by the pole on the fireman

The upward force F was 546.11 N

Explanation:

In order to find the value of the force F, you have to apply Newton's Second Law:

∑F=ma

where F represents all the forces, m is the mass and a is the acceleration.

In this case, the applied forces are the force F exerted by the pole and the weight force, due to the acceleration of gravity.

Replacing in the formula the components of the forces in the y-axis:

Using a positive direction downward:

-F + W = ma

-F + mg=ma

F = mg - ma (I)

Notice that F is upward and the weight is downward.

You need to calculate the acceleration with the given information and then you can calculate the value of F.

Applying kinematic motion formula:

Yf= Yo+ Vot + 0.5at²

where yo is the initial position, vo is the initial velocity, a is the acceleration, t is the time and Yf is the final position.

In this case:

t=1.2s

vo=0 (because the fireman starts from rest)

Yo= 0 m  and Yf= 3.0 m

Therefore:

3.0 = 0 + 0(1.2) + 0.5a(1.2)²

3.0 = 0.72a

Dividing by 0.72

a = 4.17 m/s²

Replacing it in (I):

F= 97(9.8) - 97(4.17)

F = 546.11 N

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explanation:

the answer would be   dignified - apex