Solving Masses & Friction on an Inclined Plane

[madeeeeee]

1. Homework Statement [/b]

Objects of masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The object m1 is held at rest on the floor, and m2 rests on a fixed incline of θ = 42.0°. The objects are released from rest, and m2 slides 1.50 m down the slope of the incline in 4.30 s.

a) (a) Determine the acceleration of each object. (Enter the magnitude only.)
____________- m/s2

(b) Determine the tension in the string. (Enter the magnitude only.)
______________ N

(c) Determine the coefficient of kinetic friction between m2 and the incline.
______________

my attempt
This was right
a) a=2(x)/(t)^2
a=2(1.5)/(4.3)^2
=0.16m/s^2

b) Ftension=m1(a) + m1(g)
=4(0.162)+4(9.8) = 39.848

It says this is wrong

c) u=m2(g)sin(theta)-m2(a)-(m1)(g)-(m2)(a) / (m2)(g)cos(theta)
= (9)(9.8)sin42-(4)(0.1622)-(4)(9.8)-(9)(0.1622) / (9)(9.8)cos42
=59.017-0.6488-39.2/65.55
=0.29

It says this is wrong to...

PLEASE HELP! WHAT AM I DOING WRONG!

 

[vela]

It appears your method is correct. It's probably just rounding and sig figs.

 

[grzz]

c) u=m2(g)sin(theta)-m2(a)-(m1)(g)-(m2)(a) / (m2)(g)cos(theta)
= (9)(9.8)sin42-(4)(0.1622)-(4)(9.8)-(9)(0.1622) / (9)(9.8)cos42
=59.017-0.6488-39.2/65.55
=0.29

I think that (a) and (b) are correct.
But there is a mistake in (c). The underlined part must be removed.
If a FREE BODY DIAGRAM for mass m2 is drawn, this mass 'sees' only the pull of the Earth, the tension, the frictional force and the normal reaction of the inclined plane. m2 does not 'see' the weight of m1.

 

[madeeeeee]

it still says that my b value is wrong :(

I actually wrote the formula wrong for that but I put in the right values so it was supposed to look like.
u=m2gsintheta-m1a-m1g-m2a/m2gcostheta
= (9)(9.8)sin42-(4)(0.1622)-(4)(9.8)-(9)(0.1622)/(9)(9.8)cos42
=59.0173--1.46025-39.2/65.55
=0.29

The formula should be right now but I am still getting the wrong answer

 

[vela]

Did you do the calculations carefully, taking into account significant figures? For b, you should have
$$F_T = m_1(a+g) = m_1\left(\frac{2x}{t^2}+g\right) = (4.00~\mathrm{kg})\left[\frac{2(1.50~\mathrm{m})}{(4.30~\mathrm{s})^2}+9.81~\mathrm{m/s}^2\right]$$Note that all of the quantities you're given have three significant figures, so you want to use a value of g that also has at least three significant figures. If you plug this into the calculator, you get a numerical answer like 39.88899946, which gives you a final answer of 39.9 N to three sig figs.

 

[grzz]

I think that (a) and (b) are correct.
But there is a mistake in (c). The underlined part must be removed.
If a FREE BODY DIAGRAM for mass m2 is drawn, this mass 'sees' only the pull of the Earth, the tension, the frictional force and the normal reaction of the inclined plane. m2 does not 'see' the weight of m1.

I am correcting my own post.

The underlined part is not to be removed because IT IS CORRECT. I am sorry if I caused trouble to the original poster.

The original poster did a small mistake. he did not enter (-9x0.1622) into the calculation in the next step.