Review. Three forces acting on an object are given by F → 1 = ( − 2.00 i ^ − 2.00 j ^ ) N , and F → 1 = ( − 5.00 i ^ − 3.00 j ^ ) N , and F → 1 = ( − 45.0 i ^ ) N . The object experiences an acceleration of magnitude 3.75 m/s 2 . (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s?
Review. Three forces acting on an object are given by F → 1 = ( − 2.00 i ^ − 2.00 j ^ ) N , and F → 1 = ( − 5.00 i ^ − 3.00 j ^ ) N , and F → 1 = ( − 45.0 i ^ ) N . The object experiences an acceleration of magnitude 3.75 m/s 2 . (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s?
Solution Summary: The author explains the direction of the acceleration and the forces acting on the object.
Review. Three forces acting on an object are given by
F
→
1
=
(
−
2.00
i
^
−
2.00
j
^
)
N
, and
F
→
1
=
(
−
5.00
i
^
−
3.00
j
^
)
N
, and
F
→
1
=
(
−
45.0
i
^
)
N
. The object experiences an acceleration of magnitude 3.75 m/s2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s?
(a)
Expert Solution
To determine
The direction of the acceleration.
Answer to Problem 5.22P
The direction of the acceleration is 181°.
Explanation of Solution
The forces acting on the object are F→1=(−2.00i^+2.00j^)N, F→2=(5.00i^−3.00j^)N and F→3=(−45.00i^)N. The magnitude of acceleration of the object is 3.75m/s2.
Write the formula to calculate net force act on a object
F→net=F→1+F→2+F→3
Here, F→net is the net force acting on an object, F→1, F→2 and are the given forces.
Substitute (−2.00i^+2.00j^)N for F→1, (5.00i^−3.00j^)N for F→2 and (−45.00i^)N for F→3 to find F→net.
Here, Fy is the y-component of the force and Fx is the x-component of the force.
Conclusion:
Substitute −42.0N for Fx and −1.00N for Fy in the above equation to calculate θ.
tanθ=(−1.00N−42.0N)θ=1.36°
The direction of force is equal to the direction of acceleration of object and the value of θ lies at third quadrant so the direction of acceleration with +x axis
θ=180°+1.36°=181.36°≈181°
Therefore, the direction of the acceleration is 181°.
(b)
Expert Solution
To determine
The mass of the object.
Answer to Problem 5.22P
The mass of the object is 11.2kg.
Explanation of Solution
Write the formula to calculate magnitude of net force act of an object
F=Fx2+Fy2
Write the formula to calculate mass of the object
m=Fa
Here, a is the magnitude of acceleration of an object, m is the mass of the object and F is the magnitude of net force acting on the object.
Substitute Fx2+Fy2 for F in the above equation.
m=Fx2+Fy2a
Conclusion:
Substitute −42.0N for Fx and −1.00N for Fy and 3.75m/s2 for a in the above equation to find m.
m=(−42.0N)2+(1.00)23.75m/s2=11.2kg
Therefore, the mass of the object is 11.2kg.
(c)
Expert Solution
To determine
The speed of the object after 10sec.
Answer to Problem 5.22P
The speed of the object after 10sec is 37.5m/s.
Explanation of Solution
Write the formula to calculate speed of an object
vf=vi+at
Here, vf is the final speed of an object, vi is the initial speed of an object and t is time.
Conclusion:
Substitute 0 for vi, 10sec for t and 3.75m/s2 for a to find vf.
vf=0+3.75m/s2×10sec=37.5m/s
Therefore, the speed of the object after 10sec is 37.5m/s.
(d)
Expert Solution
To determine
The velocity components of the object after 10sec.
Answer to Problem 5.22P
The x and y components of the velocity are −37.5m/s and 0.893m/s respectively.
Explanation of Solution
Write the formula to calculate velocity of an object
v→f=v→i+a→t (I)
Here, v→f is the final velocity and v→i is the initial velocity.
Write the formula to calculate mass of the object
a→=F→netm
Substitute F→netm for a→ in equation (I).
v→f=v→i+F→netmt
Substitute (−42.0i^−1.00j^)N for F→net, 11.2kg for m, 0 for v→i and 10sec for t to find v→f.
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