Force on a block

[deleted647690]

Is my reasoning correct here that the block sliding along the ledge begins to speed up due to the fact that the falling block will have constant acceleration (increasing velocity) until it reaches terminal velocity (theoretically)?
At the point it reaches terminal velocity, would the velocity of the block sliding along the ledge then be constant?

---

Members don't see this ad.

 

[aldol16]

Yes. So let's compare two states: A) the block at rest and B) the block while it's moving. When the block is at rest, the forces of static friction and tension will be equal and opposite to each other. The block remains stationary. Then let's say you tap the block so that it just barely overcomes static friction and begins moving. Now, kinetic friction governs the movement of the block. Since kinetic friction is always smaller than static friction, your tension force, which is equal to the force of static friction, will be larger than the kinetic friction force: abs(F(tension) - F(kinetic friction)) > 0. Thus, there is a net force on the block given by F(tension) - F(kinetic friction) = m*a. Since the question stem gives you that the tension and friction are constant, then their difference must also be constant such that constant = m*a. Therefore, acceleration is constant and since dv/dt = a and dv = a*dt and vf - vi = a*dt and vf = a*dt + vi. If you plot velocity as a function of time, you would thus get a straight line with slope a and y-intercept vi.

Terminal velocity is irrelevant in this case because the block would likely never reach terminal velocity.

 

[deleted647690]

Yes. So let's compare two states: A) the block at rest and B) the block while it's moving. When the block is at rest, the forces of static friction and tension will be equal and opposite to each other. The block remains stationary. Then let's say you tap the block so that it just barely overcomes static friction and begins moving. Now, kinetic friction governs the movement of the block. Since kinetic friction is always smaller than static friction, your tension force, which is equal to the force of static friction, will be larger than the kinetic friction force: abs(F(tension) - F(kinetic friction)) > 0. Thus, there is a net force on the block given by F(tension) - F(kinetic friction) = m*a. Since the question stem gives you that the tension and friction are constant, then their difference must also be constant such that constant = m*a. Therefore, acceleration is constant and since dv/dt = a and dv = a*dt and vf - vi = a*dt and vf = a*dt + vi. If you plot velocity as a function of time, you would thus get a straight line with slope a and y-intercept vi.

Terminal velocity is irrelevant in this case because the block would likely never reach terminal velocity.

Okay, so then due to the fact that there is a net force on the block that is positive, this means that acceleration must be some constant positive value. Thus, constant acceleration = increasing velocity.

 

[aldol16]

Okay, so then due to the fact that there is a net force on the block that is positive, this means that acceleration must be some constant positive value. Thus, constant acceleration = increasing velocity.

The point is that net force is not only positive but constant. This is key because this is what lets you make the jump from constant force to constant acceleration. Constant acceleration = linearly increasing velocity.