Question

a 100 kg crate is lifted vertically by a crane. the crane applies an upward force of 1200 n to lift the crate. the crate is initially at ground level and is lifted to a height of 10 meters. the force of friction opposing the lift is 200 n. calculate:

1. the net force on the crate while it is being lifted.
2. the acceleration of the crate while being lifted.
3. the gravitational potential energy of the crate after it has been lifted to 10 meters.

when writing your answers, use abbreviations for units, do not write them out (e.g. use n, do not write out newtons). correct capitalization of the units matters.
net force = 20 n
acceleration = 0.2 m/s*2
gravitational potential energy = 9800 j

Explanation:

Step1: Calculate the net force

The gravitational force \(F_g = mg\), where \(m = 100\space kg\) and \(g=9.8\space m/s^{2}\). So \(F_g=100\times9.8 = 980\space N\).
The net force \(F_{net}=F_{applied}-F_{g}-F_{friction}\).
Given \(F_{applied} = 1200\space N\), \(F_{friction}=200\space N\) and \(F_{g}=980\space N\).
\(F_{net}=1200 - 980-200\)
\(F_{net}=20\space N\)

Step2: Calculate the acceleration

According to Newton's second law \(F_{net}=ma\), where \(m = 100\space kg\) and \(F_{net}=20\space N\).
We can solve for \(a\) by \(a=\frac{F_{net}}{m}\)
\(a=\frac{20}{100}=0.2\space m/s^{2}\)

Step3: Calculate the gravitational potential energy

The formula for gravitational potential energy is \(U = mgh\), where \(m = 100\space kg\), \(g = 9.8\space m/s^{2}\) and \(h = 10\space m\)
\(U=100\times9.8\times10\)
\(U = 9800\space J\)

Answer:

net force \(= 20\space N\)
acceleration \(=0.2\space m/s^{2}\)
gravitational potential energy \(=9800\space J\)