# Pretty easy, just want someone to verify I have the right answer

• laurenbaboran
In summary, the conversation discusses a problem involving a worker pushing a box at a constant velocity with a given weight and angle of 35°. The first question asks for the value of the friction force, which is not enough information to answer. The second question asks for the acceleration of the box, but with the given information, it is not possible to determine the answer. Additional information, such as the magnitude of the applied force and the coefficient of friction, is needed to solve the problem.

#### laurenbaboran

Pretty easy, just want someone to verify I have the right answer:)

## Homework Statement

A worker pushes downward and to the left on the box at an angle of 35°.

The box weighs 135 kg and moves horizontally across the floor to the left at a constant velocity.

## Homework Equations

15. What is the value of the friction force?
a.1164 N
b.1663 N
c.2489 N
d.1324 N

16. What is the value of the acceleration of the box?
a. 6.7 m/s2
b. 0 m/s2
c. 18.4 m/s2
d. 12.3 m/s2

## The Attempt at a Solution

I worked out the numbers and arrived at 1324 for the first one and 6.7 for the second, but I would love for someone to verify this before I submit it :)

I do not think there is enough information to answer 15. For 16, what does "constant velocity" imply?

What?The angle which you gave is from horizontal or vertical?..And if the block moves horizontally across the floor to the left at a constant velocity,then how can it have acceleration?(Direction of velocity is also not changing so no production of acceleration due to change in direction of velocity)..Rephrase the question and check your answers?

You are right

voko said:
I do not think there is enough information to answer 15. For 16, what does "constant velocity" imply?

You are right.The magnitude of applied force is missing.
Just asking that if the force applied is given then shouldn't the x-component of the applied force be equal to the frictional force if the block is not accelerated?

μs=0.4
μk=o.2. My bad for leaving that out. Does that help?

Princu said:
Just asking that if the force applied is given then shouldn't the x-component of the applied force be equal to the frictional force if the block is not accelerated?

It should, but we are not given the magnitude of the applied force, nor the coefficient of friction. Knowing either one would make the problem determinate.

What is the angle relative to?

laurenbaboran said:
μs=0.4
μk=o.2. My bad for leaving that out. Does that help?
It may solve..working on it.but tell the angle

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