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Elanso [62]
3 years ago
15

A mass m1=1.5 kg rests on a 30 degree ramp with a coefficient of kinetic friction = 0.40. Mass m1 is tied to another mass m with

a string which runs over a frictionless pulley. Mass m is hanging above the ground. The acceleration of masses is measured 2.68 m/s2. What is m?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
6 0

Answer:

m = 2.31 Kg

Explanation:

given,

m₁ is the mass rest on the inclined plane = 1.5 Kg

inclination of plane = 30°

Kinetic friction = μ = 0.4

acceleration of the body = 2.68 m/s²

mass m is hanging  = ?

using equation to solve

T - m₁g sin θ - μ m₁g cos θ = m₁ a

on the block on inclined plane there will be acting tension T on the string,

mg sin θ will be the force acting opposite to the tension on the string and a frictional force will be acting which will oppose moment which will be equal to  μ m₁g cos θ

T = m₁(g sin θ +  μ g cos θ + a )

T = 1.5 (9.8 x sin 30° +  0.4 x 9.8 x cos 30°+ 2.68 )

T = 16.46 N

now, forces on the other side of pulley

m g - T = m a

m (g - a ) = T

m = \dfrac{16.46}{9.8-2.68}

m = 2.31 Kg

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m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

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m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

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Answer:

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