Speed of a 3.72 kg block pulled by cord at an angle

[Zhalfirin88]

Homework Statement

A 3.72 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 11.5 N at an angle θ = 20.5° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.070. What is the speed of the block 6.10 s after it starts moving?

Homework Equations

F=ma
v_f = at

The Attempt at a Solution

What is the normal force in this case, because it's not mg, as I tried that and got it wrong.

 

[tiny-tim]

Hi Zhalfirin88!

What is the normal force in this case, because it's not mg, as I tried that and got it wrong.

That's right … in this case, there are three forces with a (non-zero) component in the vertical direction: F N and W, instead of the usual 2.

Obviously, you need the three components to add to zero.

 

[Doc Al]

What is the normal force in this case, because it's not mg, as I tried that and got it wrong.

To figure out the normal force, add up all the vertical force components. What must they add up to?

 

[Zhalfirin88]

Hi Zhalfirin88!

That's right … in this case, there are three forces with a (non-zero) component in the vertical direction: F N and W, instead of the usual 2.

Obviously, you need the three components to add to zero.

I have no idea what you said after there are 3 non-zero components in the vertical direction.

To figure out the normal force, add up all the vertical force components. What must they add up to?

Since it's not moving in the vertical direction it'd be zero right?

 

[tiny-tim]

I have no idea what you said after there are 3 non-zero components in the vertical direction.

There's the force from the cord (F), the weight of the block, and the normal force.

Their components (in any direction) have to add to zero.

 

[Zhalfirin88]

Okay, I have no patience for this.

0 = F_N + F_g + F_T

F_N = -mg - F_T

F_N = -(3.72)(9.8) - 11.5sin(20.5)

F_N = -36.45 - 3.8523

F_N = -40.30 N

But, you plug that into static friction equation and you get f_s = -2.821

So, 11.5sin(20.5) + 2.821 = 6.8484N/3.72 kg

a = 1.841 m/s²

v_f = at

v_f = 1.841 * 6.1

v_f = 11.23 and that is wrong.

edit: I also did:

F_N = +36.45 - 3.8523 Because down is the negative direction.

F_N = 32.5977 N

But, you plug that into static friction equation and you get f_s = 2.821

So, 11.5sin(20.5) - 2.821 = 1.7455N/3.72 kg

a = .4692 m/s²

v_f = at

v_f =.4692 * 6.1

v_f = 2.86 and that is wrong. But that was my final try at the question so it's wrong for good now.

 

[Doc Al]

0 = F_N + F_g + F_T

F_N = -mg - F_T

F_N = -(3.72)(9.8) - 11.5sin(20.5)

F_N = -36.45 - 3.8523

F_N = -40.30 N

You made your mistake in your 2nd line.
ΣF = F_N + F_g + F_T
ΣF = F_N - mg + 11.5sin(20.5)

Since gravity acts down it gets a negative sign, while F_T gets a positive sign.

Since ΣF = 0:
F_N = mg - 11.5sin(20.5)