Solving 2 Physics Problems: Work and Force Calculations

[Format]

k few questions I am stuck on...

1) A student with a mass of 80.0 kg runs up three flights of stairs in 12.0 sec. The student has gone a vertical distance of 8.0 m. Determind the amount of work done by the student to elevate his body to this height. (constant speed)

and..

2) Bob is riding a rollar coster. He enconters a small hill with a radius of curviture of 12.0 m. At the crest of the hill he is lifted off his seat and held in by the safty bar. If noah is traveling with a speed of 14.0 m/s then find out the force applied by the safty bar upon Bob's 80 kg body.

Help please lol

 

[Blistering Peanut]

1) The work done is his potenial energy (mgh) divided by the time taken.
2) Newtons 3rd law, the force pulling him towards the centre is centripical force given by F = mv^2/r so the bar reacts with an equal and opposite force.

 

[AKG]

1) The work done is his potenial energy (mgh) divided by the time taken.

I don't think that's right. The work done is simply the change in energy, which in this case is strictly gravitational potential energy, mgh. What you've described (dividing by the average time) is the average power.

2) Newtons 3rd law, the force pulling him towards the centre is centripical force given by F = mv^2/r so the bar reacts with an equal and opposite force.

Right idea, but the bar doesn't react with any force, the bar is the object providing the centripetal force. By the way, who the heck is Noah? I'm guessing he meant Bob.

 

[Format]

Great, thanks guys. Yea Noah was a typo lol

 

[Gza]

lol, the guy's having an identity crisis.

 

[Format]

k sorry to bump an old post, but i still can't figure out the 2nd question here. The answer is supposed to be 507 N

 

[TALewis]

Are you sure it's 507 N? I get 522 N. Here's my solution:

At the top of the hill, two forces are acting on Bob: His weight (mg) and the normal force of the bar (F_b). They both act downward. Since Bob is undergoing uniform circular motion at the top of the hill, his acceleration is v^2/r downward. Setting up the full equation and solving for N:

[tex]
\begin{align*}
\sum F = mg + F_b &= m\frac{v^2}{r}\\
F_b &= m\frac{v^2}{r} - mg\\
\end{align}
[/tex]

Plugging in the values you gave (m = 80 kg, v = 14 m/s, r = 12 m), I get about 522 N.

 

[Format]

Yea that's what i got, i think the worksheet made Gravity = 10. Thx

 

[TALewis]

Yes, g = 10 m^2/s would give you 507 N. Did the worksheet specify that? Kind of bizarre, unless it's supposed to be overly simple.