Lesson 5 warm up problems #4

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revanthbellam
Posts: 16
Joined: Mon Aug 12, 2019 1:09 am

Lesson 5 warm up problems #4

Post by revanthbellam » Mon Jan 20, 2020 9:20 pm

Hi, maybe I just had this misconception about work the whole time, but I don't understand why the net work is 150 J as oppose to 0J for #4. I thought the formula for work was Fdcos(theta). In the problem's case, wouldn't theta be 90 degrees since the box is being lifted straight up and remaining motionless, rendering work to be 0 J? Thanks!
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

Re: Lesson 5 warm up problems #4

Post by NS_Tutor_Mathias » Mon Jan 20, 2020 9:40 pm

Theta is the angle between the force and the direction of travel. So in this case all the force is applied in the direction of travel, therefore theta = 0. So Fdcos(theta) just becomes Fd*1 = Fd ---> 100N*1.5m = 150 Nm = 150 J.

The idea is that force that can't contribute to motion because it doesn't move the object is not doing work.
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