(a)At a certain instant, a particle-like object is acted on by a force F = (4.0 N)i - (2.0 N)j + (9.0 N)k while the object's velocity is v = - (2.0 m/s)i + (4.0 m/s)k . What is the instantaneous rate at which the force does work on the object?
(b)At some other time, the velocity consists of only a j component. If the force is unchanged, and the instantaneous power is -12 W, what is the velocity of the object just then?
work=force dot dx
the dot product is scalar, so all the components (ii,jj,kk) are added.
b) power= d work/dt= d Force dot velocity/dt
-12watts= (4i-2j+9k) vj
vj= -12/-2 = 6j
To find the instantaneous rate at which the force does work on the object, we need to calculate the dot product of the force and velocity vectors. The dot product gives us the component of one vector in the direction of the other.
(a) To calculate the dot product of F and v:
F · v = (4.0 N)(-2.0 m/s) + (-2.0 N)(0 m/s) + (9.0 N)(4.0 m/s)
= -8.0 N·m/s + 0 N·m/s + 36.0 N·m/s
= 28.0 N·m/s
Therefore, the instantaneous rate at which the force does work on the object is 28.0 N·m/s.
(b) Given the instantaneous power is -12 W, we can use the definition of power to find the dot product of the force and velocity vectors.
Instantaneous power = F · v
Plugging in the given power, we have:
-12 W = (4.0 N)(v_j) + (-2.0 N)(0 m/s) + (9.0 N)(0 m/s)
Since the velocity only has a j-component, the i and k components will be zero. Simplifying the equation:
-12 W = 4.0 N(v_j)
v_j = -3 m/s
Therefore, the velocity of the object just then is -3 m/s in the j-direction.