Physics questions p3?

51) (a) Not sure about this one.

The chair should have three forces on it -- 250 N up, 160 N down (the chair) and a normal force down (due to Pat's weight and his kinematic acceleration, minus the 250 N with which he is pulling up on the rope).

For Pat, you'll have his 320 N down, the rope "pulling him up" with 250 N force, and a normal force from the chair pushing him up.

For the Pat/chair system, you've got the combined weight 480 N down and two lifting forces of 250 N each (net upward 20 N).

(b) a = net F / m = 20N / 48kg = 0.42 m/s²

using g = 10 m/s²

(c) F = m*(g+a) - T = 32kg * (9.8+0.42)m/s² - 250N

F = 77 N

71) centripetal force

Fc = mv²/r = 0.750kg * (35.0m/s)² / (60.0m*cos20.0º) = 16.3 N

for horizontal equilibrium,

Fc = T*cos20.0º + Fℓ*sin20.0º

where T is the tension and Fℓ is the lift force.

16.3 N = 0.940*T + 0.342*Fℓ

For vertical equilibrium,

Fℓ*cos20.0º = T*sin20.0º + m*g

0.940*Fℓ = 0.342*T + 0.750kg*9.8m/s²

Fℓ = 0.364*T + 7.82N

plug into horizontal

16.3 N = 0.940*T + 0.342*(0.364*T + 7.82N)

solves to

T = 12.8 N

but check the math

33) Not sure what you might be stuck on here.

Surely you can find the change in KE. (Hint: it's negative.)

Surely you can find the increase in GPE. (Hint: increase in height is distance * sin30º)

The friction force is the work done by friction (the difference in the magnitudes of the two values above) divided by the distance.

Then µ = friction force / normal force

where normal force = m*g*cosΘ