1. A 28.7 kg block (m1) is on a horizontal surface, connected to a 7.0 kg block

Question

1. A 28.7 kg block (m1) is on a horizontal surface, connected to a 7.0 kg block (m2) by a massless string as
shown below. The frictionless pulley has a radius R = 0.069 m and a moment of inertia of I = 0.110 kg m2.
A force, F = 224.1 N, acts on m1 at an angle = 28.3°. There is no friction between m1 and the surface.
Assume the blocks start from rest and there is no slipping between the pulley and the rope.
a. What is the upward acceleration of m2?
b. What is the net torque on the pulley?
c. How fast will the pulley be rotating after m1 has been pulled 75 cm?

A) the answer 2.188 m/s^2 need to see the work

B) the answer is 3.49 N*m and I need to see the work

C) The answer is 26.3 rad/s need to see the work

Explanation / Answer

Given that

A block of mass (m1) = 28.7 kg is on a horizontal surface, connected to a (m2) = 7.0 kg block by a massless string The frictionless pulley has a radius( R) = 0.069 m and a moment of inertia of (I) = 0.110 kg m2.
A force of (F1) = 224.1 N, acts on m1 at an angle ( )= 28.3°.

Let T1 and T2 are the tenisons attached in the string attached to mass ma1 andm2

Force acting on the mass m1 is

F1costheta-T1 =m1a

T1 =F1costehta-m1a

The force equation for the mass m2 is

T2-m2g =m2a

T2 =m2g+m2a

Now the net torque acting on the pulley is

(T1-T2)R =I*alpha and alpha =a/R =

solving above equations we get acceleration is

a =(F1costheta -m2g)/(I/R2+m1+m2) =(224.1)cos28.3-(7)(9.81)/( (0.110 kg m2./(0.069)2)+28.7+7)=197.314-68.67/58.804=2.1876m/s2

b)

The net torque is (t) =moment of inertia * angular acceleration =i*(a/R) =0.110 kg m2*(2.1876/0.069 m ) =3.4875N.m

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