Consider the two-body situation at the right. a 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg mass. this second mass is suspended over a pulley. the incline angle is 30.00 and the surface has a coefficient of friction 0.120. determine the acceleration of the system and the tension in the cable

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ =  30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T:  cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction  x-axis :

∑Fx = m₁*ax   ,ax  =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470  -258.77 -300*a =T

T= 1211.23-300*a   Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay   ,ay  =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T:  cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends  vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) =  Equation (2) = T

1211.23-300*a= 980  + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends  vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

there are no contact forces mentioned on the list of choices that you've provided.

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