How Is Force Calculated When Pushing a Block at Constant Velocity?

AI Thread Summary
When a car pushes a block at a constant velocity of 1 m/s, the net force on the block is zero, meaning the applied force equals the frictional force opposing it. The frictional force can be calculated using the coefficient of friction (0.22) and the weight of the block (300 kg). The relationship established is that the applied force must match the frictional force to maintain constant velocity. The discussion clarifies that while the forces are equal, the block continues to move at a steady speed. Understanding this balance of forces is crucial for solving the problem correctly.
macmac410
Messages
9
Reaction score
0

Homework Statement


a car with a constant velocity of 1m/s, accidentally hits a huge block with a mass of 300kg. Instead of stopping the driver constantly maintain the speed, pushing the block along the way until the car runs out of fuel. Find the force applied in the block
if the coefficient of friction in the block and surface is 0.22.

Homework Equations


f=ma


m=300kg
a=0 since velocity is constant

u=Ff/Fn
Fn=W=mg

u=Ff/mg
Ff=u(mg)


The Attempt at a Solution


i don't know where to start but it seems that the forced applied on the block was zero since velocity is constant? Am i riGht?
 
Physics news on Phys.org
macmac410 said:
i don't know where to start but it seems that the forced applied on the block was zero since velocity is constant? Am i riGht?
No. Since the velocity is constant the net force on the block must be zero, not the applied force. What force acts on the block opposing the applied force?
 
Doc Al said:
No. Since the velocity is constant the net force on the block must be zero, not the applied force. What force acts on the block opposing the applied force?

friction acts on the block opposing the applied force,
since you said that the net force must be zero hence!
Ffriction=Fapplied?

thanks now its getting clearer,
but how is it happened that frictional force is equal to applied force? If the two force are equal then there must be no motion on them?
 
macmac410 said:
friction acts on the block opposing the applied force,
since you said that the net force must be zero hence!
Ffriction=Fapplied?
Right. (Now you've got to figure out the friction force.)

thanks now its getting clearer,
but how is it happened that frictional force is equal to applied force? If the two force are equal then there must be no motion on them?
No, it means that the block isn't changing its velocity (once it's moving along with the car, that is). I presume that they want the applied force once constant velocity is attained.
 
Thank you very much Doc Al!
now i can find the right answer..
 
Last edited:

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging 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...
View full post »

Similar threads

Frictional Forces on a Wooden Crate with External Forces
Replies
1
Views
1K
Coefficient of Friction when Normal Force is Reduced [HS]
Replies
5
Views
2K
Newton's law problem: Pushing 2 stacked blocks on a horizontal table
Replies
13
Views
3K
Calculating Net Force and Friction: Understanding Newton's Laws of Motion
Replies
9
Views
3K
Find the force to push a book sideways on a table
Replies
2
Views
2K
Final Velocity of a car with a opposing decreasing drag force
2
Replies
57
Views
2K
How to calculate the wind force acting on a water drop?
Replies
39
Views
3K
Value of friction with constant velocity
Replies
3
Views
1K
Maximum displacement in mass spring system
Replies
5
Views
2K
Does V0 affect acceleration under constant Force?
Replies
4
Views
1K

Hot Threads

Recent Insights

Back
Top