(a) At a certain instant, a particle-like object is acted on by a force F→=(4.0N)i^−(2.0N)j^+(9.0N)k^\vec{F} = (4.0\rm{N})\hat{i} - (2.0\rm{N})\hat{j} + (9.0\rm{N})\hat{k} while the object's velocity is v→=−(2.0m/s)i^+(4.0m/s)k^\vec{v} =- (2.0\rm{m/s})\hat{i} + (4.0\rm{m/s})\hat{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 y component. If the force is unchanged and the instantaneous power is -12 W, what is the velocity of the object?

(a) the instantaneous power is [Fill in the blank],
W

(b) the velocity is [Fill in the blank],
m/s

Question

(a) At a certain instant, a particle-like object is acted on by a force F→=(4.0N)i^−(2.0N)j^+(9.0N)k^\vec{F} = (4.0\rm{N})\hat{i} - (2.0\rm{N})\hat{j} +  (9.0\rm{N})\hat{k} while the object's velocity is  v→=−(2.0m/s)i^+(4.0m/s)k^\vec{v} =- (2.0\rm{m/s})\hat{i}  +  (4.0\rm{m/s})\hat{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 y component. If the force is unchanged and the instantaneous power is -12 W, what is the velocity of the object?

(a) the instantaneous power is [Fill in the blank],
 W

(b) the velocity is [Fill in the blank],
 m/s

BlackTom AI · Accepted Answer

We start by identifying that work rate (instantaneous power) is the dot product of force and velocity: P = F · v.

(a) Given F = (4.0, -2.0, 9.0) N and v = (-2.0, 0.0, 4.0) m/s, compute the dot product:
F · v = 4.0*(-2.0) + (-2.0)*0.0 + 9.0*4.0 = -8.0 + 0.0 + 36.0 = 28.0 W.
Thus, the instantaneous power (the rate at which the force does work on the object) is 28 W.

(b) At another moment, the velocity has only a y-component: v = (0, v_y, 0). The force remains F = (4.0, -2.0, 9.0) N. The instantaneous power is P = F · v = 4.0*0 + (-2.0)*v_y + 9.0*0 = -2.0 v_y.
They tell us P = -12 W. Setting -2.0 v_y = -12 gives v_y = 6.0 m/s. Since the power is negative, the velocity component along the force’s direction is such that it yields a negative projection; here that means the object is moving in the +y direction with a magnitude of 6 m/s. Therefore, the velocity is (0, 6, 0) m/s, i.e., 6 m/s in the +y direction.

190867 Mechanics and Thermodynamics Test 2 (chapters 7-12)

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