A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Paul has magnitude 48.0 N and direction 61.0∘ south of west. How much work does Paul’s force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?
1 answer:
Answer:
-570.4 J
Explanation:
Parameters given:
Force = 48N, 61° SW
Displacement = 12m, 22.0 NE
Work done is given by:
W = F*d*cosθ
To find θ:
θ = 61 + 90 + 22 = 172°
=> W = 48 * 12 * cos172
W = - 570.4 J
Hence, the work to be done is - 570.4J.
This value is negative because the applied force is opposite the direction of the displacement.
You might be interested in
Answer:
<h2>The angular velocity just after collision is given as</h2><h2>Explanation:
As per given figure we know that there is no external torque about hinge point on the system of given mass
So here we will have
now we can say
so we will have
Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass
So we can use angular momentum conservation about the hinge point