A block with mass m = 5.00 kg slides down a surface inclined 36.9 to the horizon

Question

A block with mass m = 5.00 kg slides down a surface inclined 36.9 to the horizontal (the figure (Figure 1) ). The coefficient of kinetic friction is 0.27. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 27.0 kg and moment of inertia 0.500 kgm2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.250 m from that axis.

What is the acceleration of the block down the plane?

What is the tension in the string?

Explanation / Answer

gravitational force (Fg = m*g) is acting downward

components of Fg

Fgx = m*g*sin(theta)

Fgy = m*g*cos(theta)

Fgy = normal force N

friction force Fr = u*N

Fr = u*m*g*cos(theta)

so total force F = m*a = m*g*sin(theta) - u*m*g*cos(theta) - T     ......eq1

where u = coefficient of kinetic friction = 0.27

m = mass of block = 5 kg

r = perpendicular distance = 0.

tension is tge only force acting on the pulley which causes an angular acceleration

so torque t = l*alpha = T*r

where l = moment of inertia = 0.500 kg*m^2

alpha = angular acceleration

alpha = a/r

l*a/r = T*r

T = l*a/r^2     .....eq2

from equation 1 & 2

m*a = m*g*sin(theta) - u*m*g*cos(theta) - (I*a/r^2)

a*[m + (I/r^2)] = m*g[sin(theta) - u*cos(theta)]

a = [m*g*(sin(theta) - u*cos(theta)] / [m + (I/r^2)]

a = [5*9.81*(sin36.9 - 0.27*cos36.9)] / [5 + (0.5/(0.25)^2]

a = 18.86 / 13

a = 1.45 m/s^2

(b)

tension in string

T = l*a/r^2 = 0.50*1.45 / (0.25)^2

T = 11.6 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.