Problem #1
A 3.0 kg mass undergoes an acceleration given by a = (2.0i + 5.0j) m/s2 . Find the resultant force F and its magnitude.Answer:
Newton’s Second Law tells us that the resultant (net) force on a mass m is ∑F = ma. So here we find:
Fnet = ma = (3.0 kg)(2.0i + 5.0j) m/s2 = (6.0i + 15j)N
The magnitude of the resultant force is
Fnet = [(6.0 N)2 + (15.0 N)2]1/2 = 16.0 N
Problem #2
While two forces act on it, a particle of mass m = 3.2 kg is to move continuously with velocity (3m/s)i − (4 m/s)j. One of the forces is F1 = (2N)i+(−6N)j. What is theother force?
Answer:
Newton’s Second Law tells us that if a is the acceleration of the particle, then (as there are only two forces acting on it) we have:
F1 + F2 = ma
But here the acceleration of the particle is zero!! (Its velocity does not change.) This tells us that
F1 + F2 = 0 ⇒ F2 = −F1
and so the second force is
F2 = −F1 = (−2 N)i + (6 N)j This was a simple problem just to see if you’re paying attention!
Problem #3
A 4.0 kg object has a velocity of 3.0i m s at one instant. Eight seconds later, its velocity is (8.0i + 10.0j) m/s. Assuming the object was subject to a constant net force, find (a) the components of the force and (b) its magnitude.
Answer
(a) We are told that the (net) force acting on the mass was constant. Then we know that its acceleration was also constant, and we can use the constant–acceleration results from the previous chapter. We are given the initial and final velocities so we can compute the components of the acceleration:
ax = ∆vx/∆t = [(8.0 m/s) − (3.0 m/s)]/(8.0s) = 0.63 m/s2
and
ay = ∆vy/∆t = [(10.0 m/s) − (0.0 m/s)]/(8.0s) = 1.3 m/s2
We have the mass of the object, so from Newton’s Second Law we get the components of the force
Fx = max = (4.0kg)(0.63 m/s2) = 2.5 N
Fy = may = (4.0kg)(1.3 m/s2) = 5.0 N
(b) The magnitude of the (net) force is
Fnet = [(Fx)2 + (Fy)2]1/2 = [(2.5 N)2 + (5.0 N)2]1/2 = 5.6 N
and its direction θ is given by
tan θ = Fy/Fx = 5.0/2.5 = 2.0 ⇒ θ = tan−1(2.0) = 63.40
(The question didn’t ask for the direction but there it is anyway!)
Problem #4
Five forces pull on the 4.0kg box in Fig.(1). Find the box’s acceleration (a) in unit–vector notation and (b) as a magnitude and direction.
Answer:
(a) Newton’s Second Law will give the box’s acceleration, but we must first find the sum of the forces on the box. Adding the x components of the forces gives:
∑Fx = −11 N + 14 N cos 30◦ + 3 .0 N = 4.1 N
(two of the forces have only y components). Adding the y components of the forces gives:
∑Fy = +5.0 N + 14 N sin 300 – 17 N = −5.0 N
So the net force on the box (in unit-vector notation) is
∑F = (4 .1 N)i + (−5.0 N)j. Then we find the x and y components of the box’s acceleration using a = ∑F/m:
ax = ∑Fx/m = (4.1 N)/(4.0 kg) = 1.0 m/s2
ay = ∑Fy/m = (−5.0N)/(4.0kg) = −1.2 m/s2
So in unit–vector form, the acceleration of the box is
a = (1.0 m/s2)i + (−1.2 m/s2)j
(b) The acceleration found in part (a) has a magnitude of
a = [ax2 + ay2]1/2 = [(1.0 m/s2)2 + (−1.2 m/s2)2]1/2 = 1.6 m/s2
and to find its direction θ we calculate
tan θ = ay/ax = −1.2/1.0 = −1.2
which gives us:
θ = tan−1(−1.2) = −500. Here, since ay is negative and ax is positive, this choice for θ lies in the proper quadrant.
So the net force on the box (in unit-vector notation) is
∑F = (4 .1 N)i + (−5.0 N)j. Then we find the x and y components of the box’s acceleration using a = ∑F/m:
ax = ∑Fx/m = (4.1 N)/(4.0 kg) = 1.0 m/s2
ay = ∑Fy/m = (−5.0N)/(4.0kg) = −1.2 m/s2
So in unit–vector form, the acceleration of the box is
a = (1.0 m/s2)i + (−1.2 m/s2)j
(b) The acceleration found in part (a) has a magnitude of
a = [ax2 + ay2]1/2 = [(1.0 m/s2)2 + (−1.2 m/s2)2]1/2 = 1.6 m/s2
and to find its direction θ we calculate
tan θ = ay/ax = −1.2/1.0 = −1.2
which gives us:
θ = tan−1(−1.2) = −500. Here, since ay is negative and ax is positive, this choice for θ lies in the proper quadrant.
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