Solving Homework: Man Pushes 11.2 kg Box

In summary, a man is pushing a 11.2 kg box on a horizontal surface at a constant speed of 3.5 m/s2. The coefficient of friction is 0.20. To maintain this motion, the man must apply a horizontal force of 61.152 N, calculated by using the equations F=ma and F=μN. If the force is removed, the box will slide a distance of 21.95 meters before coming to rest.
  • #1
tatiana
43
1

Homework Statement


Man pushes a box that is 11.2 kg on a horizontal surface at a constant speed of 3.5 m/s2. Where μ is .20
a)What is the horizontal force rthat teh man must apply to maintain teh motion
b)If the force of part a is removed, how far does the box slide before coming to rest?

Homework Equations


F=ma
F=μN


The Attempt at a Solution


F=ma
F=11.2(3.5)
F=39.2

then
F=μN
F=(.20)(109.76)... [11.5*9.8]
F=21.95

then to get the total force i added 39.2+21.95 and got a total of 61.152

For part b I would asume that without part a the force would just be 21.95 because it doesn't account for the answerr to part a?
 
Physics news on Phys.org
  • #2
Man pushes a box that is 11.2 kg on a horizontal surface at a constant speed of 3.5 m/s2

Constant speed or acceleration?
 

FAQ: Solving Homework: Man Pushes 11.2 kg Box

How do I determine the force needed to push an 11.2 kg box?

To determine the force needed to push an 11.2 kg box, you will need to use the formula F = ma, where F is the force, m is the mass of the box (11.2 kg), and a is the acceleration (which is equal to 0 since the box is not moving). So the force needed to push the box is 0 Newtons.

What if the box is on a ramp or incline?

If the box is on a ramp or incline, the force needed will depend on the angle of the incline and the coefficient of friction between the box and the surface it is on. You can use the formula F = mg(sinθ + μcosθ), where F is the force, m is the mass of the box, g is the acceleration due to gravity (9.8 m/s²), θ is the angle of the incline, and μ is the coefficient of friction. This will give you the minimum force needed to push the box up the incline without it sliding back down.

Can I use a lower force to push the box?

Yes, you can use a lower force to push the box. The force needed to push the box will depend on the weight of the box, the surface it is on, and the angle of the incline (if applicable). In some cases, you may need to use more force to overcome static friction and get the box moving, but once it is moving, you can use less force to keep it in motion.

How can I increase the force needed to push the box?

To increase the force needed to push the box, you can either increase the weight of the box or increase the angle of the incline (if applicable). You can also decrease the coefficient of friction between the box and the surface it is on. Additionally, applying a force at an angle (instead of straight on) can also increase the overall force needed.

Are there any other factors that can affect the force needed to push the box?

Yes, there are other factors that can affect the force needed to push the box. These can include the surface roughness, air resistance, and any external forces acting on the box (such as wind). The force needed can also change if the box is on a moving surface, like a conveyor belt.

Similar threads

Stockroom worker pushes a box homework problem
Replies
2
Views
4K
Friction problem (pushing a box)
Replies
3
Views
3K
Newton's Third Law: Acceleration of box and worker
Replies
2
Views
4K
Magnitude of acceleration of boxes -- is this method correct?
Replies
7
Views
3K
How Is the Friction Force on the Box Calculated?
Replies
2
Views
4K
Man pulling a cube on a rough surface
Replies
2
Views
1K
Find force box exerts on other box up incline
Replies
11
Views
2K
Newton's Second law: Pushing the box
Replies
6
Views
2K
Efficiency of Ramp: Solving Box Push Up Problem
Replies
3
Views
3K
Basic Newton's Laws and Applied Force: 3 boxes
Replies
5
Views
8K

Hot Threads

Recent Insights

Back
Top