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work - mu a student athlete holding 100 lbs over their head a student d…

Question

work - mu
a student athlete holding 100 lbs over their head
a student dragging their chair across the floor
a student lifting a backpack off the floor
a teacher moving a desk across the room

Explanation:

Step1: Recall the work formula

Work \(W = F\times d\times\cos\theta\), where \(F\) is force, \(d\) is displacement, and \(\theta\) is the angle between force and displacement. If \(d = 0\) or \(\cos\theta=0\), \(W = 0\).

Step2: Analyze each case

  • A student athlete holding 100 lbs over their head: Here, \(d = 0\) (no displacement of the weight in the direction of the force exerted by the athlete). So, using \(W=F\times d\times\cos\theta\) with \(d = 0\), \(W = 0\). So, it is “No Work”.
  • A student dragging their chair across the floor: There is a force applied in the direction of the displacement of the chair (\(\theta = 0^{\circ},\cos\theta = 1\)). So, \(W=F\times d\times\cos\theta=F\times d\times1

eq0\). So, it is “Work Done”.

  • A student lifting a backpack off the floor: The force applied (upward) and the displacement (upward) have \(\theta = 0^{\circ},\cos\theta = 1\). Using \(W = F\times d\times\cos\theta\), \(W

eq0\). So, it is “Work Done”.

  • A teacher moving a desk across the room: The force applied (assuming in the direction of motion) and displacement have \(\theta = 0^{\circ},\cos\theta = 1\). Using \(W=F\times d\times\cos\theta\), \(W

eq 0\). So, it is “Work Done”.

Answer:

A student athlete holding 100 lbs over their head: No Work
A student dragging their chair across the floor: Work Done
A student lifting a backpack off the floor: Work Done
A teacher moving a desk across the room: Work Done