Question

1. determine the force that a 65.0 kg man exerts on the bottom of an

a. at rest.
b. accelerating up at 4.90 m/s².
c. moving at constant velocity down.
d. moving down at 1.23 m/s².

Explanation:

Step1: Analyze forces at rest

When at rest, by Newton's second law \(F = ma\), \(a = 0\). The force \(F_N\) (normal force, which is the force on the bottom) equals the weight \(mg\). \(m = 65.0\space kg\), \(g=9.8\space m/s^{2}\).
\(F_N=mg\)
\(F_N = 65.0\times9.8\)

Step2: Analyze when accelerating up

Using \(F_N - mg=ma\) (from \(F_{net}=ma\), \(F_{net}=F_N - mg\) as \(F_N\) is upward and \(mg\) is downward).
\(F_N=m(g + a)\)
\(m = 65.0\space kg\), \(g = 9.8\space m/s^{2}\), \(a=4.90\space m/s^{2}\)
\(F_N=65.0\times(9.8 + 4.90)\)

Step3: Analyze when moving at constant velocity

When \(v=\) constant, \(a = 0\). Again, \(F_N=mg\) (similar to step 1)
\(F_N=65.0\times9.8\)

Step4: Analyze when moving down with acceleration

Using \(mg - F_N=ma\) (\(F_{net}=mg - F_N\) as \(mg\) is downward and \(F_N\) is upward). Then \(F_N=m(g - a)\)
\(m = 65.0\space kg\), \(g = 9.8\space m/s^{2}\), \(a = 1.23\space m/s^{2}\)
\(F_N=65.0\times(9.8- 1.23)\)

Answer:

a. \(F_N=65.0\times9.8 = 637\space N\)
b. \(F_N=65.0\times(9.8 + 4.90)=65.0\times14.7=955.5\space N\)
c. \(F_N=65.0\times9.8 = 637\space N\)
d. \(F_N=65.0\times(9.8 - 1.23)=65.0\times8.57 = 557.05\space N\)