Mass of block given μK and Force?

[Ace.]

Homework Statement

A student pushes a block with 240 N of force across a horizontal surface with a coefficient of kinetic friction of 0.4. The block accelerates at a rate of 0.88 m/s². Find the mass of the block.

Homework Equations

μ_K = F_K / F_N
F=ma

The Attempt at a Solution

My first attempt was to figure out mass using F=ma

m = f/a

= 240 N / 0.88 m/s²​

= 272.7 kg​

but this is not taking into account the kinetic friction...

So what I did was

F_K = μ_K x F_N

= 0.4 x (272.7 kg x 9.8 m/s²​

= 1069.1 N​

I know I'm doing this wrong and I need to know how to derive at net force to find acceleration

 

[Doc Al]

My first attempt was to figure out mass using F=ma

m = f/a

= 240 N / 0.88 m/s²​

= 272.7 kg​

but this is not taking into account the kinetic friction...

To use ƩF = ma you must include all the forces.

So what I did was

F_K = μ_K x F_N

= 0.4 x (272.7 kg x 9.8 m/s²​

= 1069.1 N​

You are using the incorrect answer from your first attempt.

Instead, just call the mass "m" and set up your equation using the net force. Then you can solve for the mass. What's the friction force in terms of "m"?

 

[Ace.]

You are using the incorrect answer from your first attempt.

Instead, just call the mass "m" and set up your equation using the net force. Then you can solve for the mass. What's the friction force in terms of "m"?

μ_K = F_K / F_N
F_N = F_K / μ_K
mg = F_K / μ_K
m = (F_K / μ_K) / g

Like this?

 

[Doc Al]

μ_K = F_K / F_N
F_N = F_K / μ_K
mg = F_K / μ_K
m = (F_K / μ_K) / g

Like this?

No. (That won't work, since F_K is proportional to the mass, so your equation ends up being m = m.)

Use ƩF = ma. First work on the forces. There are two forces acting. What are they?

 

[Ace.]

force applied of 240 N and kinetic friction

I can't find magnitude of kinetic friction because I cannot find Force normal.

 

[Doc Al]

force applied of 240 N and kinetic friction

Good.

I can't find magnitude of kinetic friction because I cannot find Force normal.

If the mass is "m", what's the weight?

 

[Ace.]

Good.If the mass is "m", what's the weight?

weight = 9.8m

 

[Doc Al]

weight = 9.8m

Good. So now you have the normal force. Keep going.

 

[Ace.]

μ_K = F_K / F_N
F_K = μ_K × F_N
F_K= 0.4 x 9.8m

I have 2 unknowns?

 

[Doc Al]

μ_K = F_K / F_N
F_K = μ_K × F_N
F_K= 0.4 x 9.8m

Good.

I have 2 unknowns?

The only unknown is the mass, which is what you are trying to find.

Keep going. What is ƩF?

 

[Ace.]

F_applied = 0.88m


just guessing here:
Fnet = F_A [fwd] + F_f [bwd]
Fnet = F_A [fwd] - F_f [fwd]
Fnet = 0.88m - (0.4 x 9.8m)

 

[Doc Al]

F_applied = 0.88m

The applied force is given as 240 N.

just guessing here:
Fnet = F_A [fwd] + F_f [bwd]
Fnet = F_A [fwd] - F_f [fwd]

This is good.

Fnet = 0.88m - (0.4 x 9.8m)

Fix this and then use Fnet = ma.

 

[Ace.]

oh wow my bad...

Fnet = 240 - (0.4 x 9.8m)

= 240 - 2.92m​

Still don't get what I need to do next?

 

[Doc Al]

oh wow my bad...

Fnet = 240 - (0.4 x 9.8m)

= 240 - 2.92m​

Still don't get what I need to do next?

Apply Newton's 2nd law: Fnet = ma

 

[Ace.]

240 - 2.92m = 0.88m
240 = 0.88m + 2.92m
240 = 3.8m
m = 63 kg

 

[Doc Al]

240 - 2.92m = 0.88m
240 = 0.88m + 2.92m
240 = 3.8m
m = 63 kg

That's the right idea. But I think you made a mistake earlier (probably a typo):

Fnet = 240 - (0.4 x 9.8m)

= 240 - 2.92m​

Fix that, resolve your equation, and you'll have the correct mass.

 

[Ace.]

Fnet = 240 - (0.4 x 9.8m)
= 240 - 3.92m

240 - 3.92m = 0.88m
240 = 0.88m + 3.92m
240 = 4.8m
m = 50 kg

wohooooooo thanks a lot!

 

[Doc Al]

Yay!