Problem 2.5 (15 points) Calculate the acceleration of the lkg box that slides do

Question

Problem 2.5 (15 points) Calculate the acceleration of the lkg box that slides down the 300 inclined plane. The coefficient of friction between the box and the plane is 0.1. Problem 2.6 (15 points) A 50kg box rests on the elevator's floor. Elevator moves up so that its velocity changes from 8m/s to 2m/s in 3 seconds. Find the apparent weight of the box Problem 2.7 (15 points) See the picture below. mi-5kg: m-10kg: plane is inclined at 30 to horizontal, coefficient of friction = 0.1 . Calculate the acceleration and direction of the motion of the 10kg block. m2 mi Problem 2.8 (15 points) A 100g sample is spun in the centrifuge at 10cm distance from the spinning center with 1000g acceleration. Calculate the centrifuge rotation speed in RPM.

Explanation / Answer

Problem 2.5

m*g*sin(theta) - u*m*g*cos(theta) = m*a

a=9.81*(sin(30) - 0.1*cos(30))

a=4.05 m/s^2

Problem 2.6

Now a=(2-8)/3 = -2 m/s^2

N - m*g = m*a

N = m*( g + a )

= 50*( 9.81 - 2 )

= 390.5 N

Problem 2.7

From figure, let T be the tension in the rope

m2*g - T = m2*a

98.1 - T = 10*a

T + 10*a = 98.1 ----- (1)

also T - m1*g*sin(30) - u*m1*g*cos(30) = m1*a

T - 5*a = -28.77 -------(2)

Solving (1) and (2), we get

T = 13.52 N and a=8.458 m/s^2 downwards

Problem 2.8

a = w^28r

w = sqrt( 1000*9.81 / 0.1 )

=313.2 rad/s

= 2990.92 RPM ans

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