# rasen58

Senior Members

78

1. ## Velocity from parabolic force time graph

Okay thank you for the corrections on all those points!
2. ## Kinetic friction on smooth then rough surface

Okay thanks, I guess the textbook gave the wrong answer.
3. ## Kinetic friction on smooth then rough surface

Given a 2.0 kg mass at rest on a horizontal surface at point zero. For 30.0 m, a constant horizontal force of 6 N is applied to the mass. For the first 15 m, the surface is frictionless. For the second 15 m, there is friction between the surface and the mass. The 6 N force continues but the mass slows to rest at the end of the 30 m. The coefficient of friction between the surface and the mass is _____. To solve this, I found the velocity at 15 m by first using F=ma to find that the acceleration for the first 15 m is 3 m/s^2. Then I used a kinematic equation to find that the velocity at 15 m is sqrt(90). So then, for the second 15 m, I drew a force diagram and saw that for the mass to decelerate to 0 in the exact same distance as it took to accelerate, then the net force must be the same magnitude but in the reverse direction to slow it down. So I thought that since it was previously just 6 N to the right, I thought the force of friction would have to be 12 N to the left so that the net force is 6 N to the left. Force of friction = coeff_fric * m * g So that means that 12 = coeff_fric * m * g Solving for coeff_fric, I got 0.6. But that is apparently wrong since it's supposed to be 0.3. But the only way to get 0.3 is if Force of friction = 6 N to the left. But I don't see why it should be 6 N instead of 12 N to the left.
4. ## Velocity from parabolic force time graph

Find the speed of a 3 kg mass when it reaches the 10 m mark. The net force is in the same direction as the motion. Moving on straight line (rectilinear) I think I'd have to use net work = change in kinetic energy theorem here. Since the net work would be the area under the force position graph, I counted the boxes under the graph. So I tried just counting all the boxes and got (around?) 39 boxes. I then multipled that by 20 since each box is 20 N and got 780 N Since W = F *d = 780 N * 10 m = 7800 N*m So 7800 = .5 m v^2 Solving for v, I got 22.8 m/s. But the answer is actually 24. I think that's due to me not counting the correct number of boxes? or is there a different way or a better way to approximate how many boxes are under the graph? Thanks
5. ## Thermal energy and temperature

There are two equal masses of silver insulated from each other. Sample A is maintained at 20 degrees C. Sample B has twice as much thermal energy as Sample A. Silver has a specific heat of 233 J/g*C. What is the temperature of sample B? I don't have any idea how to do this and don't have that good of a grasp of the term 'thermal energy'. I think thermal energy is equivalent to total kinetic energy of all the particles, so maybe you could use the formulas for average kinetic energy, but I'm not sure how. q = cm dT KEavg = (1/2)mv2 = (3/2)KT
6. ## How much ice melts at equilibrium

Oh wow that makes sense. Thanks! So it would still be at 0 degrees C.
7. ## How much ice melts at equilibrium

So if the net heat is 0 q = heat gained by raising 3kg ice from -10oC to 0oC + heat gained by converting 3kg ice to water at 0oC + heat gained raising 3kg of water fom 0oC to +xoC degrees + Heat lost by lowering 10kg water from +10oC to +xoC = 0 q = (1/2)(3)(0 - (-10)) + (3)(80) + (1)(3)(x-0) + (1)(10)(x-10) = 0 But after I solved for x now, I get 11.9 degrees C, which doesn't make sense either since it's greater than 10.
8. ## How much ice melts at equilibrium

I'm not sure why that's wrong.
9. ## How much ice melts at equilibrium

3 kg of ice whose temperature is -10 C is dropped into 10 kg of water whose temperature is 10 C. How much of the ice melts when the mixture is brought to equilibrium? I wanted to find the equilibrium temperature first so I used q = cm T And c (water) = 1 kcal/kg C, c (ice) = 0.5 kcal/kg C So (0.5)(3)(x - (-10)) = (1)(10)(10 - x) That turns into 1.5x + 15 = 100 - 10x x = 7.4 C So I then determined how much energy outputted when lowering the temperature of the water is q = cm T = (1)(10)(10 - 7.4) = 26 kcal Then, to bring the ice to 0 C requires q = cm T = (0.5)(3)(0 - (-10)) = 15 kcal Then, to cause the phase change, I found the latent heat. Heat of fusion = 80 kcal/kg q = mL = (3 kg)(80 kcal/kg) = 240 kcal But 240 is so much higher than 26, so how is this possible. What am I doing wrong?

11. ## Force required to remove a stopper from a flask

A flask is closed tightly with a rubber stopper 3.00 cm in diameter. All the air in the flask is then removed. How much force is necessary to remove the stopper from the evacuated flask? (1 atm = 1.013 x 10^5 N/m^2) I have no idea how to do this problem. I think I'm supposed to use Bernoulli's equation, but I'm not sure how exactly to use it. I also tried using P = F/A So F = P*A = 1 x 10^5 x pi x (0.03)^2 = 283 But the answer is 71.6 N.
12. ## Ideal gas law problem

Oh thanks, forgot about that, it works now
13. ## Ideal gas law problem

When tripling the volume of a given weight of gas at 27 C, keeping the pressure constant, the temperature must be raised to? I used V1/T1 = V2/T2 and set V1 to 1 L 1/27 = 3/T2 T2 = 81 C. But the answer says it's 627 C. Is it wrong?
14. ## Constant quantities in circular motion

Yep, just learned that it is perpendicular so it makes sense now.
15. ## Constant quantities in circular motion

A child runs in a circular path with constant speed. Which one of the following quantities is constant and non-zero? A. Linear velocity B. Angular velocity C. Centripetal acceleration D. Angular acceleration E. Total acceleration I thought none of them would be constant since the direction of the vectors would always be changing when you move in a circle. But since I had to pick an answer, I thought it might be between linear velocity and angular velocity and picked linear velocity since there is a constant speed. But the answer is angular velocity. Why?
16. ## Balance a meter stick

Thank you so much for all the help once again!
17. ## Balance a meter stick

Okay so I finally understand how to actually go about doing these problems now. So I made the fulcrum the pivot point and found the torques from there. So d is the length from the fulcrum to the left end 100-d is the length from the fulcrum to the right end So counter-clockwise moment is (.0024 kg)(10 m/s^2)(d/2) Clockwise moments are (.0024 kg)(10 m/s^2) ( (100-d)/2 ) & (2.16 kg)(100 -d) So the net torque = (.012)(d) - [ 1.2 - (.012)(d) ] - [ 216 - 2.16(d) ] = 0 Then I solved for d and got 99.45, which is off. What did I do wrong this time? Edit: So I did it again with the pivot point being the right end. So net torque = (Fbeam)(.5 m) - (Ffulcrum)(x) = 0 where x is the distance from the 100 cm to the fulcrum. So Ffulcrum has to balance the weight of the beam and the mass so it turns out to be 24 N. And then I solved for x and got .05 m, which means it's at the 95 cm mark.
18. ## Balance a meter stick

I thought it would pivot on the fulcrum. It was like that, but I don't understand why there are three forces.
19. ## Balance a meter stick

Yes, I drew a diagram. Why do you need the centre of gravity though?
20. ## Balance a meter stick

Oh, I see, it makes sense that it would be 0.024 kg per centimetre. But why is the counterclockwise moment (0.024gd) x(d/2) N-cm? Where did the d/2 come from? I got the mass on the right side is (100-d)(.024 kg), but how do I calculate the moment?
21. ## Balance a meter stick

A uniform meter stick of mass 0.24 kg can be balance by a mass of 2.16 kg at the 100 cm mark if the fulcrum is placed at the point in cm ___. So the mass on one side is just the mass of the stick (0.24 kg) and the other side is (2.16 kg + .24 kg = 2.4 kg) So the torques have to be equal So on one side, it's 0.24 kg * 10 m/s^2 * d And the other is 2.4 kg * 10 m/s^2 * (100-d) So I set them equal to each other and got 26.4 d = 2400 d = 90 cm. But the answer is apparently 95 cm. why?
22. ## Pulley friction dissipation

As shown in the figure, a 2kg mass and a 3kg mass hanging from the massless pulley are released from rest. After the 3 kg mass has fallen 1.5 m, it is moving with a speed of 3.8 m/s. What is the rate at which the frictional force dissipates energy during this time interval? My solution: The 3 kg mass has Fg down and a Tension force up and the 2kg mass has a tension force to the right and a friction force to the left. With my coordinate system, the tension forces cancelled out, and I got Fnet = Fg - Ffr Ffr = Fg - Fnet Fg = 3 * 10 = 30 N Fnet = Mtotal * a To find the acceleration of the system, I knew Vi = 0m/s, Vf = 3.8 m/s, y = 1.5m Vf2 = Vi2+ 2ay Then by solving for a, I got 4.813 m/s^2 So now I know Fnet = (5 kg)*4.813 = 24.065 N And Ffr = 30 N - 24.065 N = about 6 N To find the dissipation of energy, power = work/time work = force*displacement The force here is the friction force = 6 N And I have to find displacement and time now. So to find time, looking at the 3 kg block y = Vit + 1/2 at2 Then solving for t gave me .7895 seconds. So in .7895 seconds, the 3kg block fell 1.5 meters, so now I need to find how much the 2kg block moved in that time period. So x = Vit + 1/2 at2 I use v initial = 0 m/s again (right?) and the same a that I found along with the t = .7895 And got 1.5 meters. So now to find the power dissipated, (6 N * 1.5 meters)/.7895 s = 11.39 Watts But the actual answer is 10 Watts. What did I do wrong? Thanks.
23. ## Block at half its final speed

Thanks v2 = 2ax x = L So then 1/2 of that final v = 1/2 sqrt(2aL) And the equation for distance is now just x = 1/2 at2 So we need to find a or t So a = (v-v0)/t and I then simplified that into t2=1/2 L/a And plugged that back into the distance equation to get 1/4 L. I think that's right?
24. ## Block at half its final speed

A block of mass M starting from rest slides down a frictionless inclined plane of length L. When the block has attained 1/2 its final speed, the distance it has traveled along the plane is choices: L/4, L/2, L/sqrt(2), 3L/4 I tried using the equation V2=v02+2ax and set v0 = 1/2v which becomes 1/4 v2 to get 3/4 v2 = 2ax and then 3a/8 v2 = x But I don't think I'm doing it right. What should I be doing?
25. ## Satellite orbital time without any numbers

Oh okay, that worked for me, just didn't realize I should've used the earth constants. Thanks.
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