Question

Newton's Second Law question, half way done just need some guidance!

Answer 1

You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle 30 degree above the horizontal with a force of magnitude 240 N, causing the box to move horizontally. The force of friction between the moving box and the floor has magnitude 41.5 N . What is the box's acceleration just after it begins to move?

Answer 2

So I found the x acceleration:
applied force F=240N at an angle of 30degrees is = F cos 30 = 240 * 0.866 = 207.846 N
force of friction = 41.5 N
Normal force = 207.846 - 41.5 = 166.346 N
F= m * a and a = F / m = 166.346 / 30 = - 5.54487 m/s2

Answer 3

Just want to make sure it's correct and how to find the y component.

Answer 4

yes method for finding acceleration is correct ( so is the answer).
Finding y component you need to multiply the vector (in this case 240 N) by sin(30)

Answer 5

also is the negative correct? it should - 5.54487 m/s2 yes?

Answer 6

and y component is (240sin30-41.5)/30 ?

Answer 7

@ProfBrainstorm please help if you can!

Answer 8

i don't get why you are taking negative...
The negative only describes the direction, but in this case the acceleration is towards that force applied.
Therefore the acceleration should be positive.

Force of y component is 240 x sin(30)
Net force of y component is (240sin(30) - 0)

I have taken zero because the friction force is only horizontal and not vertical. As there seems to be no apparent opposing force, therefore it is zero.

Mass is a Scalar quantity and therefore will remain same for both horizontal and vertical components.
Therefore acceleration of y component is (240sin(30))/30.

Answer 9

You only have acceleration in the x (horizontal) direction, there is no need to consider vertical components of anything in this problem.
The acceleration of the box will be given by (240 x cos30 -41.5)/30

Answer 10

for this problem it specifically asks: Find both the x and y components of the acceleration of the box.
Enter the x and y components of the acceleration in meters per second squared, separated by commas.

Answer 11

@ProfBrainstorm I'm so silly, you're right it's just 0!

Answer 12

@ProfBrainstorm one last Q, sorryy for the bother:
A normal walking speed is around 2.0 m/s . How much time t does it take the box to reach this speed if it has the acceleration 5.54m/s2 that you calculated above?

Answer 13

I got 0.36s

Answer 14

yes, 0.36s would be correct.