Question 3 of 10 UNIVERSITY PHYSICS Philip R. KestenDavid L. Tauck presanted by

Question

Question 3 of 10 UNIVERSITY PHYSICS Philip R. KestenDavid L. Tauck presanted by Map sapling leaming Two blocks are connected over a massless, frictionless pulley as shown in the figure below. The mass of block 2 is 11.7 kg, and the coefficient of kinetic friction between block 2 and the incline is 0.200. The angle of the incline is 25.5°. If block 2 is moving up the incline at constant speed, what is the mass of block 1? Number 2.11 Friction There is additional feedback available! View this feedback by clicking on the bottom divider bar. Click on the divider bar again to hide the additional feedback Incorrect. Clase Previous Give Up & View Solution # Try Again Next Exit Explanation

Explanation / Answer

The goal here is to identify the forces and put them into equations that ultimately involve only values we are given or already know, such as g = 9.81m/s^2.

(1) Fnet = Fup - Fdown

From Newton:

(2) Fnet = (m1 + m2) * a

Let's analyze the right side of (1): Fup is generated by the force of gravity g on m1:

(3) Fup = m1 * g

and Fdown comprises two forces for which we can write more detailed equations:

(4) Fdown = Fg2 + Ff

[NOTE: The process from here is to break down the right side of (4) so that you are using some of the information given, such as , the angle of the incline, and k, the coefficient of sliding friction. When you do that, you can solve for m2, which is 0.572kg.]

We know what the force of gravity on m2 is--its weight W2; we are interested in the component of W2 is parallel to the incline:

(5) Fg2 = W2 * sin

The force of friction Ff is given by:

(6) Ff = Fn * k,

where Fn, the normal force, is the component of W2 that is perpendicular to the incline:

(7) Fn = W2 * cos

and

(8) W2 = m2 * g


Now, substituting (7) into (6):

(6.1) Ff = Fn * k= (W2 * cos) * k

And we can substitute into (4) from (5) and (6.1):

(4.1) Fdown = Fg2 + Ff

= (W2 * sin) + (W2 * cos * k) and (8)

= m2 * g * (sin + cos * k)


Now, taking (1) and substituting (2) for the left side and (3) and (4.1) for the right:

(9) (m1 + m2) * a = Fup - Fdown

= (m1 * g) - (m2 * g * (sin + cos * k) )

Expanding and factoring (9) successively:

(9.1) m1 * a + m2 * a = (m1 * g) - (m2 * g * (sin + cos * k) )

(9.2) m2 * a + m2 * g * (sin + cos * k) = m1 * g - m1 * a

(9.3) m2 * (a + g * (sin + cos * k) ) = m1 * (g - a)

Solving for m2:

(10) m2 = m1 * (g - a) / (a + g * (sin + cos * k) )

= 11.7 * (9.81 - 0) / (0 + 9.81 * (sin(25.5) + cos(25.5) * 0.200) )

= 114.777/2.61

m2=43.97 kg

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.