PHYS 201 SD2 SD3 Solution S2026

Ajouté : 2026-05-13

[00:00:00]Okay, so the solutions for today's quizzes, we'll start with the easy ones.

[00:00:04]SD3A, you have a car driving around the corner, radius 60 meters, it's raining, so the coefficient of static friction is 045. What's the maximum speed at which the car can make the turn around the corner? So remember from what we did in chapter 8, the car has a normal force F grab. Stack friction keeps the tires on the road and the net force has to be some triple force pointed towards the center of the turn. the curve. So, Fnet equals static friction. Fnet is centrial. So, that's always MV^2 over R when you're moving in a circle.

[00:00:38]Stagic friction is mu time the normal force. And we know that the normal force is just mg when you're just sitting flat on a surface. So, mass cancels. We just do algebra. V= square root mus * rg. The 045 * the 60 has 9.8. The velocity speed is 16.27. 27 m/s.

[00:01:01]Okay, SD3B was the same problem, uh, different numbers and.35 * 36 m * 9. M/s squared. That velocity was 11.11 m/s.

[00:01:15]Okay, SC2A. So, this is just like what we did in class. You have two masses, a heavier block m1, a lighter block m2 is being pulled at 30 newtons. There however is a net acceleration of 1.5 m/s squared. So it's not steady anymore. We have to worry about fnet. Uh 7 kilos is 1 kilo. What's the tension in the rope right here? And what is mu k? So we'll start with m2 cuz that's the easiest free body diagram to do. So with m2 you have tension going up and gravity going down. So fnet equals tension minus m2g.

[00:01:49]So fnet is m2 net equals tension minus m2g. So the tension we solve for that is m2a net plus m2g. So it's a 1 kilogram mass. We have the net acceleration here, gravity here. So the total tension is 11.3 newtons.

[00:02:09]What happens with m1? We got three forces to worry about. We've got f of pulling to the right. We got tension pulling to the left. And we have fk friction also pulling to the left. So that equals Fnet. We're solving for FK because we're trying to go after mu k.

[00:02:26]So fk= f - t minus fnet. F is 30 newtons. The tension we found over here.

[00:02:33]Fnet is m1 * a net. So that's the 7 kg * the 1.5 m/s squared. So the total friction force is 8.2 me 8.2 newtons.

[00:02:43]Once you have that, we know that FK equals mu k m1g. And we're solving for muk. 8.2 / the mass* 9.8. 8 muk is 0.12.

[00:02:54]All right, SD2B, same problem, different numbers. Uh, net acceleration is 1 m/s squared. Um, and we have a 6 kg M1. Same procedure for M2, the easy one. Fnet equals tension minus the weight. M2 net equals tension minus M2G. Tension is 10.8 Newtons. For M1, Fnet is force minus tension minus friction. Do the math here. Friction force becomes 8 18.2 Newtons. MUK equals.31 when you work it all out. All right. So, you have any questions about this, please hit me up.