Objects of masses m 1 = 4.00 kg and m 2 = 9.00 kg are connected by a light strin

Question

Objects of masses

m1 = 4.00 kg

and

m2 = 9.00 kg

are connected by a light string that passes over a frictionless pulley as in the figure below. The object

m1

is held at rest on the floor, and

m2

rests on a fixed incline of

= 43.0°.

The objects are released from rest, and

m2

slides 1.55 m down the slope of the incline in 4.35 s.

(a) Determine the acceleration of each object. (Enter the magnitude only.)
  m/s2

(b) Determine the tension in the string. (Enter the magnitude only.)
N

(c) Determine the coefficient of kinetic friction between

m2

and the incline.

Explanation / Answer

part a )

s = 1/2*a*t^2

a = 2s/t^2

a = 2 * 1.55 / (4.35)^2

a = 0.164 m/s^2

part b )

T - m1g = m1a

T = m1(g+a)

T = 39.895 = 39.9 N

part c )

for m2:

sum fo force in y direction : Fy = n - m2gcos43 = 0 (n is the normal force)

sum of forces in x direction : Fx = m2gsin43 - fk - T = m2a

(where fk is the kinetic friction, T is tension of the string)

for m1:

Fy = T - m1g= m1a

Step 2:

Combine these equations to cancel Tension:

m2gsin43 - fk - T = m2a

T - m1g= m1a

And we get:

m2gsin43 - fk -T + T -m1g = m2a + m1a

m2gsin43 - fk -m1g = m2a + m1a

m2gsin43 - mu* (m2g cos43) -m1g = a(m2+ m1)

(9kg)(9.8m/s^2)(sin43) - mu*(9kg)(9.8m/s^2)(cos43) - (4kg)(9.8m/s^2) = 0.164 m/s^2 (9kg+4kg)

20.95 - mu* 64.51N = 2.132 N

mu = 0.292

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