A 82.0 kg firefighter slides down a pole while a constant frictional force of 30

Question

A 82.0 kg firefighter slides down a pole while a constant frictional force of 300 N retards her motion. A horizontal 20.0 kg platform is supported by a spring at the bottom of the pole to cushion the fall. The firefighter starts from rest 3.25 m above the platform, and the spring constant is 4000 N/m.

(a) Find the firefighter's speed just before she collides with the platform.
m/s

(b) Find the maximum distance the spring is compressed. (Assume the frictional force acts during the entire motion.)
m

Explanation / Answer

A)
aceeleration of the person = g - Friction/m

= 9.8 - 300/82

= 6.14 m/s^2

Apply, v^2 - u^2 = 2*a*s

v = sqrt(2*a*s) (here initial speed, u = 0)

= sqrt(2*6.14*3.25)

= 6.32 m/s <<<<<<<<<<<<<<<<<-----------------------------Answer

B) let y is the maximum compression.

Apply energy conservation

0.5*m1*v1^2 + (m1+m2)*g*y - Friction*y   = 0.5*k*x^2

0.5*82*6.14^2 + (82+20)*9.8*y - 300*y = 0.5*4000*y^2

1546 + 1000*y - 300*y = 2000*y^2

2000*y^2 - 700*y - 1546 = 0

solving the above equation we get

y = 1.07 m <<<<<<<<<<<<<<<<<-----------------------------Answer

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