Question

T Corolla

How do I work out how much energy is required to move 12.5kg in a circular motion 180 degrees. It takes 2 sec to preform this action and the radius is 32cm total diastance is 202cm I cannot find formula. thanks folks

bnt

That's because you're missing some information - or you haven't told us everything. Here's where I have to make some assumptions:

• the system is all horizontal movement - no gravity is involved;
• the motion is in two parts: speeding the block up, then slowing it to a standstill, as efficiently as possible.;
• the two parts are identical: the acceleration occurs in half the time, over half the distance;
• acceleration and deceleration are linear.

With those stated assumptions, I'd look for a way of calculating the velocity of the block at the halfway point (90 degrees), given that it has to get there in one second. With the velocity you can work out the kinetic energy of the block at that point, which is the work done to accelerate it. Then you'd do the same amount of work in stopping it, so multiply that by 2.

I could give you a formula for the velocity, but if this is a homework assignment, you'll be expected to show where that formula came from, so I think it's better for you to figure it out. Hint: basic calculus.

The answer I got:

max v = 1.005 m/sec, total Work done = 25.266 J

T Corolla

bnt wrote: »

That's because you're missing some information - or you haven't told us everything. Here's where I have to make some assumptions:

• the system is all horizontal movement - no gravity is involved;
• the motion is in two parts: speeding the block up, then slowing it to a standstill, as efficiently as possible.;
• the two parts are identical: the acceleration occurs in half the time, over half the distance;
• acceleration and deceleration are linear.

With those stated assumptions, I'd look for a way of calculating the velocity of the block at the halfway point (90 degrees), given that it has to get there in one second. With the velocity you can work out the kinetic energy of the block at that point, which is the work done to accelerate it. Then you'd do the same amount of work in stopping it, so multiply that by 2.

I could give you a formula for the velocity, but if this is a homework assignment, you'll be expected to show where that formula came from, so I think it's better for you to figure it out. Hint: basic calculus.

The answer I got:

max v = 1.005 m/sec, total Work done = 25.266 J

Thanks for reading my post. I trying to work out how muck energy(J) I use to complete 15 reps if barbell curl. so I can convert it to calories burned thanks again

bnt

Ah - but with barbell curls you're dealing with vertical motion, against gravity, which means that the answers I gave aren't complete. See what I mean about giving a complete description of the problem? What you wrote read like some university Dynamics assignment, so I treated it as one.

I think you can get a simplified answer by just adding in the energy used to raise the barbell, using the formula [latex]W = m\;g\;h[/latex], where m is the mass, g is the constant of gravitational acceleration. h is the vertical displacement in metres, which in this case is 2x the radius. So the extra W is 12.5 x 9.81 x 0.64 = 78.48 J , and the total is just under 104.75 J.

Since 1 kCal is 4184 J, you'd need to do about 40 reps to burn 1 kCal, though (of course) there's a lot more going on, when you exercise, than just moving weights around. (A Snickers bar in Ireland contains about 270 kCal of food energy!) Calculations like this assume perfect efficiency (not gonna happen) and don't cover the aerobic effects, heat, and so on. For another view, have a read of this and the follow-up reply.

PS: it occurs to me that that figure might look a little low, but it's not a lot of weight mass, nor is it being moved over such a long distance. You weigh much more than 12.5kg, and you regularly move yourself over much greater vertical distances e.g. when you climb stairs. Curls might feel hard because the force is concentrated on a few specific muscles, but the picture is more complicated than that e.g. read this about the need for both aerobic and weight training. :pac:

T Corolla

bnt wrote: »

Ah - but with barbell curls you're dealing with vertical motion, against gravity, which means that the answers I gave aren't complete. See what I mean about giving a complete description of the problem? What you wrote read like some university Dynamics assignment, so I treated it as one.

I think you can get a simplified answer by just adding in the energy used to raise the barbell, using the formula [latex]W = m\;g\;h[/latex], where m is the mass, g is the constant of gravitational acceleration. h is the vertical displacement in metres, which in this case is 2x the radius. So the extra W is 12.5 x 9.81 x 0.64 = 78.48 J , and the total is just under 104.75 J.

Since 1 kCal is 4184 J, you'd need to do about 40 reps to burn 1 kCal, though (of course) there's a lot more going on, when you exercise, than just moving weights around. (A Snickers bar in Ireland contains about 270 kCal of food energy!) Calculations like this assume perfect efficiency (not gonna happen) and don't cover the aerobic effects, heat, and so on. For another view, have a read of this and the follow-up reply.

PS: it occurs to me that that figure might look a little low, but it's not a lot of weight mass, nor is it being moved over such a long distance. You weigh much more than 12.5kg, and you regularly move yourself over much greater vertical distances e.g. when you climb stairs. Curls might feel hard because the force is concentrated on a few specific muscles, but the picture is more complicated than that e.g. read this about the need for both aerobic and weight training. :pac:

Thanks again I will post on the fitness form and see can I get an idea of what I am burning off