Question

a 5.0 kg object is moving forward with an applied force of 10.0 n right, displacing 5.0 m. the total work done due to the force of gravity on the object is: 50 j 250 j 500 j 5000 j 0 j which of the following is equivalent to the units of a joule (j)? n/m n/kg kgm/s nkg kg*m^2/s^2

Explanation:

First question:

Step1: Recall the work - formula

The work formula is \(W = F\cdot d\cdot\cos\theta\), where \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and the displacement.
For the force of gravity, \(F = mg\) (acting vertically downwards). The object is moving horizontally. So, \(\theta=90^{\circ}\) and \(\cos\theta = 0\).

Step2: Calculate the work done by gravity

Substitute into the work formula \(W = F\cdot d\cdot\cos\theta\). Since \(\cos90^{\circ}=0\), \(W = 0\) (regardless of the values of \(F\) and \(d\)).

Second question:

Step1: Recall the formula for work

Work \(W = F\cdot d\). Force \(F=ma\) (Newton's second law \(F = ma\), where \(m\) is in \(kg\) and \(a\) is in \(m/s^{2}\), so \(F\) has units \(kg\cdot m/s^{2}\)). Displacement \(d\) has units \(m\).

Step2: Calculate the units of work

\(W=(kg\cdot m/s^{2})\cdot m=kg\cdot m^{2}/s^{2}\)

Answer:

1. \(0J\)
2. \(kg\cdot m^{2}/s^{2}\)