A boy is pushing a chair by applying a force of 5 newtons. His mother helps him push it faster by applying an additional force o
2 answers:
Answer: Option (d) is the correct answer.
Explanation:
Net force means the sum of total or all the forces acting on an object.
Mathematically, ....and so on
Therefore, when a boy is pushing a chair by applying a force of 5 newtons and his mother is helping him by pushing the chair with a force of 7 newtons then calculate the net force acting on the chair as follows.
= (5 + 7) N
= 12 N
Thus, we can conclude that net force acting on the chair is 12 N.
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Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force =
Equating both, we have
So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
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
Answer:
Explanation:
If a particle move with time and expressed according to the formula:
f(t) = 0.01t⁴ − 0.03t³
a) Velocity is the change in motion of the particle with respect to time and it is expressed as;
Hence the velocity of the particle at time t is
b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:
Hence the velocity after 1second is -0.05
c) The particle is at rest when when the time is zero.
Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second