A block with mass m 1=2.10 kg rests on a frictionless table. It is connected wit

Question

A block with mass m1=2.10 kg rests on a frictionless table. It is connected with a light string over a pulley to a hanging block of mass m2

=4.10 kg. The pulley is a uniform disk with a radius of 3.10 cm and a mass of 0.470 kg (the figure below).

1) Calculate the acceleration of each block. (Express your answer to three significant figures.)

m/s2

2) Calculate the tension in the x -direction (horizontal part of string). (Express your answer to three significant figures.)

N

3)Calculate the tension in the y -direction (vertical part of string). (Express your answer to three significant figures.)

N

4) How long does it take the blocks to move a distance of 2.25 m? (Express your answer to three significant figures.)

s

5) What is the angular speed of the pulley at this time? (Express your answer to three significant figures.)

rad/s

Explanation / Answer

let,

mass of the block, m1=2.1 kg and m2=4.1 kg

mass of the pulley, m3=0.47 kg and radius, r=3.1cm

a)

in case of mass m1,

m1*a=T1 -----(1)

and

in case of mass m2,

m2*a=m2*g-T2

T2=m2*(g-a) -----(2)

and

in case of pulley,

Torque=I*alpa

(T2-T1)*r=1/2*m3*r^2*(a/r)

(T2-T1)=1/2*m3*(a) ----(3)

(m2*(g-a)-m1*a)=1/2*m3*a

(4.1*(9.8-a)-2.1*a)=1/2*0.47*a

=====> a=6.24 m/sec^2

acceleration, a=6.24 m/sec^2

b)

from equation no (1) ,

T1=m1*a

T1=2.1*6.24

T1=13.1 N

c)

from equation no (2),

T2=m2*(g-a)

T2=4.1*(9.8-6.24)

T2=14.6 N

d)

d=u*t+1/2*a*t^2

2.25=0+1/2*6.24*t^2

====> t=0.849 sec

e)

angular speed , w=alpa*t

w=(a/r)*t

w=(6.24/(3.1*10^-2))*0.849

===> w=170.9 rad/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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