How do Newton's Laws work in these situations?

AI Thread Summary
In the discussion, participants analyze the application of Newton's Laws to various scenarios involving friction and forces. For a 5kg box pulled by a 20N force, the friction force equals 20N, indicating the box does not move due to static friction being greater than the applied force. In the sled scenario, the correct acceleration is calculated as 2.99 m/s², with some confusion about rounding and force components. Additionally, participants clarify the definitions of kinetic and static friction coefficients, emphasizing the importance of using static friction when the object is not moving. Understanding these concepts is crucial for accurately solving physics problems involving forces and motion.
vu10758
Messages
95
Reaction score
0
1) A 5kg box is pulled by a force F horizontally (no angle). There is friction between the box and horizontal surface. Mu_k = .25 and Mu_s = .5. If F=20N find the magnitude of the friction force and the acceleration of the block.

I used the formula |f_k| = mu_k * |Normal force|

f_k = .25 * 9.8 * 5
f_k = 12.25 N

The correct answer is that the friction is 20 N, meaning that the block didn't move. Where did I go wrong?

2) A 5kg sled is pulled an an angle 5 degree above the horizontal. Friction between the sled and the snow is described by mu_k = .1 and mu_s = .15. What is acceleration of the sled?

The correct answer is 2.99 m/s^2

This is what I did.

I took the 20 N and multiply by cos (5 degree) to get 19.92.

|f_k| = mu_k * N
f_k = .1*5*9.8
f_k = 4.9

(19.92 - 4.9) / 5 = a
a = 3.004

Did I just round my answer differently or did I do something wrong?

3) A 2kg object is under the influences of several forces. Assume that the net force is constant. At t=0, the object's velocity is 3i + 2j and at t=4s its velocity is 11i - 14j . Find the magnitude and direction of net force acting on the object. Please explain this to me. I have no idea how to start.

Thanks in advance for your help.
 
Physics news on Phys.org
Can you please tell me what Mu_k, Mu_s and |f_k| all mean? I'm doing similar stuff to this at the moment but I don't recognise those things you are using.
 
Mu_k is the coefficient of kinetic friction
Mu_s is coefficient of static friction

|f_k| is kinetic friction
|f_s| is static friction
 
Oh okay gotcha. Yeah I actually probably can't help because I haven't yet delved into static and kinetic friction really. I have worked with simply friction before without determining differences between the two
 
Before the box moves, it is static friction that is important. You calculated the acceleration using kinetic friction. Go back and do it using static friction.
 
Thanks for your help. Since the static friction is greater than the force pushing the object, the object will not move. Since 20N is applied on the object, the friction is also 20N according to Newton's Third Law. Am I correct?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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...
Back
Top