Before a collision, the x-momentum of an object is 8.0 × 103 kilogram meters/second, and its y-momentum is 1.2 × 104 kilogram me
2 answers:
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Answer:
a) Fₓ = 23.5 N
b) Net force = Fₓ
Explanation:
An image of the question as described is attached to this solution.
From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)
a) From the diagram, the horizontal component of the force is
Fₓ = 25 cos 20° = 23.49 N = 25 N
b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain
Net force = Fₓ - Frictional force
But frictional force is 0 N
Net force = Fₓ
Hope this Helps!!!
Answer:
v₀ =3.8 m/s
Explanation:
Newton's second law of the box:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Known data
m=2.1 kg mass of the box
d= 5.4m length of the roof
θ = 20° angle θ of the roof with respect to the horizontal direction
μk= 0.51 : coefficient of kinetic friction between the box and the roof
g = 9.8 m/s² : acceleration due to gravity
Forces acting on the box
We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.
W: Weight of the box : In vertical direction
N : Normal force : perpendicular to the direction the roof
fk : Friction force: parallel to the direction to the roof
Calculated of the weight of the box
W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N
x-y weight components
Wx= Wsin θ= (20.58)*sin(20)° =7.039 N
Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N
Calculated of the Normal force
∑Fy = m*ay ay = 0
N-Wy= 0
N=Wy =19.34 N
Calculated of the Friction force:
fk=μk*N= 0.51* 19.34 N = 9.86 N
We apply the formula (1) to calculated acceleration of the block:
∑Fx = m*ax , ax= a : acceleration of the block
Wx-f = ( 2.1)*a
7.039 - 9.86 = ( 2.1)*a
-2.821 = ( 2.1)*a
a=(-2.821) /( 2.1)
a= -1.34 m/s²
Kinematics of the box
Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement = 5.4 m
v₀: initial speed
vf: final speed = 0
a : acceleration of the box = -1.34 m/s²
We replace data in the formula (2)
0²=v₀²+2*(-1.34)*(5.4)
2*(1.34)*(5.4)= v₀²
v₀ = 3.8 m/s