Physics Homework Help - please check my work?

Question

Hi, I need help with the following problem:
Starting from rest, a 90-kg firefighter slides down a fire pole. The average frictional force everted on him by the pole has a magnitude of 820 N, and his speed at the bottom of the pole is 3.9 m/s. How far did he slide down the pole?

So here is what I get from the problem:
mass(m) = 90 kg
Initial velocity (Vo)= 0 m/s
Force of static friciton (Fs)= 820 N
Final Velocity (Vf) = 3.9 m/s

Need to find :
height (ho)

So I set up my equation:
Work of nonconservative force (Wnc) = (1/2mVf^2 + mghf) - (1/2 mVo^2 + mg(ho)

So using the following equation for work,
work of frictions(Wf)= F(cos θ) s, I get
Wf= Fs (cos 180)(ho)
Wf = -(Fs)(ho)

My total initial energy is:
Ei = mg(ho)

My toal final energy is:
Ef = 1/2 mV^2

So I get,
-Fs(ho) = 1/2 mV^2 - mg(ho)

Here is where I get confused since I can't set the equation to solve for ho. Any help will be greatly appreciated.

Thanks!

Answer 1

"Here is where I get confused since I can't set the equation to solve for ho."

Why not? It's not the way I would have done it, but it should be valid.



-Fs(ho) = 1/2 mV^2 - mg(ho)

mg(ho) - Fs(ho) = (1/2) mV^2

(mg -Fs)(ho) = (1/2) mV^2

(ho) = (1/2)mV^2 / (mg -Fs)

(ho) = (1/2) 90kg (3.9m/s)^2 / (882N - 820N)

(ho) = 684J / 62N

(ho) = 11.0m



Alternate solution:

v^2 = 2ax

x = (1/2) v^2 / a

x = (1/2) v^2 / (F(net)/m)

x = (1/2) v^2 / ((mg - Fs) / m)

x = .5 (3.9m/s)^2 / ((882N - 820N) / 90kg)

x = 7.605 m^2/s^2 / (62N / 90kg)

x = 7.605 m^2/s^2 / 0.689 m/s^2

x = 11.0m

Answer 2

This part is basic algebra, once you see through the confusion of so many symbols. The key is to recognize which symbols are unknown variables and which symbols are known. Fs, m, and V are given, and g is a constant. Only ho is unknown. So your equation has this form:



Ax = B - Cx



Unless you have totally forgotten algebra, you know how to solve that. Get all terms with x on one side, then use distributive law, then divide.



Ax + Cx = B

(A + C)x = B

x = B / (A + C)



Using the actual symbols:



-Fs(ho) = (1/2) mV^2 - mg(ho)

-Fs(ho) + mg(ho) = (1/2) mV^2

(-Fs + mg)(ho) = (1/2) mV^2

ho = (1/2) mV^2 / (-Fs + mg)



Plug and chug.

Answer 3

By F(net) = F(gravity) - F(friction)

=>ma = mg - Ff

=>90a = 90 x 9.8 - 820

=>a = 0.69 m/s^2

By v^2 = u^2 + 2as

=>(3.9)^2 = 0 + 2 x 0.69 x h

=>h = 11.04 m

Answer 4

(90 x 9.8) = 882N., acting down.

Friction applies 820N. upwards.

Net force = 62N. acting down.

Acceleration down pole = (f/m) = 62/90, = 0.688(r) m/sec^2.

Distance slid down pole = (v^2/2a), = (3.9^2/1.377(r), = 11.04 metres.