Answer:
324.3Nm
Explanation:
The torque is given by the equation
[tex]\tau=r\ X\ F[/tex]
in this case the vectors r and F are perpendicular between them, thus:
[tex]\tau=rF[/tex]
The forces acting on the mass are:
[tex]T-Mg=Ma[/tex] (1)
where T is the tension of the cable, M is the mass and a is the acceleration.
Furthermore, we have that the acceleration is:
[tex]a=\frac{v-v_0}{t}=\frac{10m/s-0m/s}{10s}=1\frac{m}{s^2}[/tex]
By replacing in (1) we can obtain:
[tex]T=Ma-Mg=M(a-g)=(150kg)(1\frac{m}{s^2}+9.8\fac{m}{s^2})=1620N[/tex]
The force T produces the torque on the pulley, hence:
[tex]\tau=rT=(0.2m)(1620N)=324Nm[/tex]
the answer is 324.4Nm
hope this helps!
