Respuesta :
Answer:
a = -.51 m/s²
coefficient of friction = .05
fk= 12.25 N
Explanation:
mass = 20 + 5kg
weight = 25kg ·9.8 = 245 N
There are many ways to do this I am sure but I did this:
1. found a by using vf²=vi²-2aΔx. accel is - because it is slowing down.
started at 4.5 m/s, final = 0 and Δx=20....a=-.51 m/s²
2. the force acting on the object is the friction force which has to equal the mass(accel).
-fk (going to left on free body diagram) = ma
-mu·n = ma
-mu·245N = 25 (-.51)
mu=.05
fk= .05(245N) = 12.25 N
The kinematics and Newton's second law allow to find the results for the questions of the movement of the child and the skateboard are:
a) the deceleration is a = 0.51 m / s²
b) The friction force is: fr = 12.75 N
c) The friction coefficient is equal to: μ = 0.052
Given parameters
- The child's mass M = 20 kg.
- The mass of the skateboard m = 5 kg.
- Initial velocity vo = 4.5 m / s.
- The distance traveled x = 20 m.
To find
a) Acceleration.
b) The friction force.
c) The coefficient of friction.
a) Kinematics studies the movement of bodies, finding relationships between their position, speed and acceleration.
Since the child ends up at rest, his final velocity is zero (v = 0), let's use the relation.
v² = v₀² - 2 a x
0 = v₀² - 2ax
a = [tex]\frac{v_o^2}{2x}[/tex]
a = [tex]\frac{4.5^2 }{2 \ 20}[/tex]
a = 0.51 m / s²
b) Newton's second law states that the net force is proportional to the product of the mass and the acceleration of the body.
A free-body diagram in a diagram of the forces without the details of the body, in the attached we can see the free-body diagram of the system.
Let's write Newton's second law, since the boy and the scooter do not separate, we use the total mass.
y-axis
N-W = 0
N = W
N = (M + m) g
N = (20 +5) 9.8
N = 245 N
x-axis
fr = [tex]m_{total} \ a[/tex]
Fr = (20+ 5) 0.51
fr = 12.75 N
c) Friction force is a macroscopic force of the interactions between the two surfaces and has the formula.
fr = μ N
μ = [tex]\frac{fr}{N}[/tex]
μ = 12.75 / 245
μ = 0.052
In conclusion, using kinematics and Newton's second law we can find the results for the questions of the movement of the boy and the skateboard are:
a) The deceleration is a = 0.51 m / s²
b) The friction force is: fr = 12.75 N
c) The friction coefficient is equal to: μ = 0.052
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