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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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       The radius of the uniform disk is R_d = 0.200m

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Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

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Substituting values

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               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

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    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

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The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

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                  = \frac{0.868^2}{0.200^2}

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The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

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                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

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