What horizontal force is required to accelerate the block

Click For Summary
SUMMARY

The discussion focuses on calculating the horizontal force required to accelerate a 4.0 kg block at 5.0 m/s² on a surface with a coefficient of kinetic friction of 0.25. The net force needed for acceleration is determined using the equation ΣF = ma, resulting in a required net force of 20 N. To overcome friction, the frictional force is calculated as Fk = μ * N, yielding 9.8 N. The total force required to achieve the desired acceleration while overcoming friction is therefore 29.8 N.

PREREQUISITES
  • Understanding of Newton's Second Law (ΣF = ma)
  • Knowledge of frictional forces and coefficients (kinetic friction)
  • Ability to construct and interpret free body diagrams
  • Basic algebra for solving equations
NEXT STEPS
  • Study the effects of different coefficients of friction on force calculations
  • Learn about free body diagram construction for complex systems
  • Explore advanced applications of Newton's laws in dynamic systems
  • Investigate the relationship between mass, acceleration, and force in various contexts
USEFUL FOR

Students in physics, engineers working on mechanical systems, and anyone interested in understanding the principles of force and motion in practical applications.

a7med2009
Messages
8
Reaction score
0
A block of mass 4.0kg rests on a horizontal surface. What horizontal force is required to accelerate the block at 5.0m/s2 if the coefficient of kinetic friction is 0.25?
 
Physics news on Phys.org


start with writing a free body diagram.
& any equations you may know that you think would be useful here.
 


ΣF=ma

Fk=mu * N
 


ma=4.0kg*5.0m/s^2
=20 N
Fk= 0.25 * 20
= 5 NIs it correct?
 


so, how much force do you need to apply in order to overcome the frictional force?
 


ΣF=F-fk=mg
F=mg-fk
=(4.0)*(9.8)-(0.25)(39.2)
=29.4 N
 


In your last post, you've got a vertical force (mg) and a horizontal force (fk) in the same equation. Not allowed.

In your previous post, you found the frictional force's maximum value. You also found that in order to accelerate at 5m/s/s, you need 20 Newtons of NET force.

So how much TOTAL force must be applied (horizontally) to overcome friction and give the block enough force to accelerate?
 
Last edited:


ohh 0_o

it's
ΣF=F-fk=ma
F=ma+fk
=(4.0)*(5)+(0.25)(39.2)
=20+9.8
=29.8 N
 


looks correct to me
 

Similar threads

Newton's law problem: Pushing 2 stacked blocks on a horizontal table
  • · Replies 13 ·
Replies
13
Views
3K
Force pulling on a string at an angle with a block at the other end
  • · Replies 2 ·
Replies
2
Views
1K
Friction problem of a block sandwiched between 2 surfaces
  • · Replies 6 ·
Replies
6
Views
1K
Find acceleration of Moving incline with a block on it
  • · Replies 11 ·
Replies
11
Views
3K
Calculating Force and Acceleration in a Two Block System with Varying Friction
  • · Replies 6 ·
Replies
6
Views
1K
Normal force acting on a block on an accelerating wedge
  • · Replies 12 ·
Replies
12
Views
3K
Can a 10N Force Overcome Friction to Move a 10kg Block?
  • · Replies 2 ·
Replies
2
Views
2K
Find the acceleration of the system (2 blocks sliding on a table)
  • · Replies 23 ·
Replies
23
Views
2K
Calculating acceleration of a prism and block connected to a wall through a rope and pulley
  • · Replies 4 ·
Replies
4
Views
868
Accelerating wedge with a block at rest on it
  • · Replies 43 ·
2
Replies
43
Views
4K