A 68.0 kg firefighter slides down a pole while a constant frictional force of 30

Question

A 68.0 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 4.10 m above the platform, and the spring constant is 4000 N/m.

A 68.0 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 he firefighter starts from rest 4.10 m above the platform and the spring constant is 40 N/m (a) Find the firefighter's speed just before she collides with the platform 6.65 m/s (b) Find the maximum distance the spring is compressed. (Assume the frictional force acts during the entire motion.) Note, WebAssign has you solve this problem as if the platform was not compressed before the collision. 0.867 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error m

Explanation / Answer

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

= 9.8 - 300/68

= 5.39 m/s^2

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

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

= sqrt(2*5.39*4.10)

= 6.65 m/s^2. Ans

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*68*5.39^2 + (68+20)*9.8*y - 300*y = 0.5*4000*y^2

987.77 + 862.4*y - 300*y = 2000*y^2

2000*y^2 - 562.4*y - 987.77 = 0

solving the above equation we get

y = 8.67 m

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