Question

1. what is the force on a persons hand, which is using a rope to accelerate a 5 kg block upward with an acceleration of 2.2 m/s²?

Explanation:

Step1: Apply Newton's second law

Newton's second law is \(F = ma\). Here, the mass \(m = 5\space kg\) and the acceleration \(a=2.2\space m/s^{2}\).

Step2: Calculate the net force

The net force \(F_{net}=ma\). Substitute \(m = 5\space kg\) and \(a = 2.2\space m/s^{2}\) into the formula: \(F_{net}=5\times2.2\).

Step3: Consider the force due to gravity

The force due to gravity \(F_g=mg\), where \(g = 9.8\space m/s^{2}\) and \(m = 5\space kg\), so \(F_g=5\times9.8\).

Step4: Find the force on the hand

The force on the hand \(F\) (tension in the rope) is given by \(F - F_g=F_{net}\). Then \(F=F_{net}+F_g\). Substitute \(F_{net}=5\times2.2\) and \(F_g = 5\times9.8\) into the equation: \(F=5\times(2.2 + 9.8)\).

Answer:

\(F = 60\space N\)