A particle is to move in an xy plane, clockwise around the origin as seen from the positive side of the z axis. In unit-vector n
otation, what torque acts on the particle if the magnitude of its angular momentum about the origin is (a) 4.0 kg ⋅ m2/s, (b) 4.0t2 kg ⋅ m2/s, (c) 4.0 √ t kg ⋅ m 2 / s and (d) 4.0/t2 kg ⋅ m2/s?
1 answer:
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Answer:
-4.0 N
Explanation:
Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):
(1)
We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be
And we can fidn the acceleration by using the formula:
where
v = 0 is the final velocity
u = 1.75 m/s is the initial velocity
t = 2.25 s is the time the box needs to stop
Substituting, we find
(the acceleration is negative since it is opposite to the motion, so it is a deceleration)
Therefore, substituting into eq.(1) we find the force of friction:
Where the negative sign means the direction of the force is opposite to the motion of the box.