# Answer to Question #713 in Mechanics | Relativity for needra

Question #713

Two masses of 2.25 kg each, connected by a string slide down a ramp making an agnle of 32 degrees with the horizontal/ the coeffcient of kinetic fricyion between m1 and the ramp is 0.46. The coefficient of kinetic friction between m2 and the ramp is 0.19. Find the magnitude of the acceleration of the masses. What is the tension in the string?

Expert's answer

We have to solve the following system of equations:

mg sin(32) - mg*mu1*cos(32)+T=ma;

mg sin(32) - mg*mu2*cos(32)-T=ma.

2mgsin(32) - g(mu1+mu2)m*cos(32)=2ma;

a= g (2 sin(32) - (mu1+mu2) cos(32) )/2 = 2.54 m/s^2

T= m(a+g* mu1*cos(32)-g sin(32)) = 2.25 (2.54+ 10*0.49*cos(32)-10*sin(32)) = 3.14 N

Answers: acceleration would be 2.54 m/s^2, the tension in the sting - 3.14 N.

mg sin(32) - mg*mu1*cos(32)+T=ma;

mg sin(32) - mg*mu2*cos(32)-T=ma.

2mgsin(32) - g(mu1+mu2)m*cos(32)=2ma;

a= g (2 sin(32) - (mu1+mu2) cos(32) )/2 = 2.54 m/s^2

T= m(a+g* mu1*cos(32)-g sin(32)) = 2.25 (2.54+ 10*0.49*cos(32)-10*sin(32)) = 3.14 N

Answers: acceleration would be 2.54 m/s^2, the tension in the sting - 3.14 N.

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## Comments

Assignment Expert02.10.10, 14:05D-densiity, mass=4.8kg, vol=5.5L

D=mass/vol

D=(4.8 kg/5.5L)*(1000g/1kg)*(1L/1000mL)=0.87g/mL

Answer:0.87g/mL

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