Calculating work for a weightlifter.

  • Thread starter Thread starter tbarnet7
  • Start date Start date
  • Tags Tags
    Work
AI Thread Summary
The discussion centers on calculating the work done by a weightlifter lifting 390 N weights over a distance of 1.75 m. The formula used is W=F∆x, leading to an initial calculation of 682.5 Joules. Participants confirm the calculation, with one member initially doubting their result but ultimately agreeing with the original answer. The consensus is that the work done is indeed 682.5 Joules. The thread highlights the importance of double-checking calculations in physics.
tbarnet7
Messages
3
Reaction score
0

Homework Statement


A weight lifter lifts a set of 390 N weights from ground level to a position over his head, a vertical distance of 1.75m. How much work does the weight lifter do, assuming he moves the weights at a constant speed?

I am not confident in my physics skills so I am really looking for someone to grade my work and give me feedback; thanks!

Homework Equations


W=F∆x


The Attempt at a Solution


W=(390N)(1.75m)=682.5 Joules
 
Physics news on Phys.org
W=(390N)(1.75m) looks good!
I get a slightly different answer than your 682.5; better run it through the calculator again.
 
I would like to point out that I get the same value as OP.
 
My mistake; thanks for catching it.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

B Why does the work done by a weightlifter result in a net energy loss?
Replies
24
Views
3K
Calculating Tension Using Work Energy Theorem
Replies
1
Views
2K
Work power and energy by a weight lifter
Replies
3
Views
7K
Finding change in internal energy
Replies
2
Views
4K
Finding Equilibrium in a Weightlifting System: Torque and Statics Explained
Replies
17
Views
3K
To calculate the forces on a composite body
Replies
2
Views
1K
Minimum force to tip the bin over
Replies
7
Views
922
Virtual Work to find equilibrium condition
Replies
1
Views
2K
Three part weightlifter question
Replies
2
Views
3K
Finding the WORK done by a crane in lifting a load
Replies
14
Views
8K

Hot Threads

Recent Insights

Back
Top