Respuesta :
Answer:
[tex]\Delta t =1.31\ s[/tex]
Explanation:
given,
coefficient of kinetic friction, μ = 0.25
Speed of sled at point A = 8.6 m/s
Speed of sled at point B = 5.4 m/s
time taken to travel from point A to B.
we know,
J = F Δ t
J is the impulse
where F is the frictional force.
t is the time.
we also know that impulse is equal to change in momentum.
[tex]J = m(v_f - v_i)[/tex]
frictional force
F = μ N
where as N is the normal force
now,
[tex]F\Delta t = m(v_f -v_i)[/tex]
[tex]\mu m g \times \Delta t = m(v_f-v_i)[/tex]
[tex]\mu g \times \Delta t = v_f-v_i[/tex]
[tex]\Delta t =\dfrac{v_f-v_i}{\mu g}[/tex]
[tex]\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}[/tex]
[tex]\Delta t =1.31\ s[/tex]
time taken to move from A to B is equal to 1.31 s
Answer:
Time taken by the sled is 1.31 s
Solution:
As per the question:
Coefficient of kinetic friction, [tex]\mu_{k} = 0.25[/tex]
Velocity at point A, [tex]v_{A} = 8.6\ m/s[/tex]
Velocity at point A, [tex]v_{B} = 5.4\ m/s[/tex]
Now,
To calculate the time taken by the sled to travel from A to B:
According to the impulse-momentum theorem, impulse and the change in the momentum of an object are equal:
Impulse, I = Change in momentum of the sled, [tex]\Delta p[/tex] (1)
[tex]I = Ft[/tex] (2)
where,
F = Force
t = time
p = momentum of the sled
Force on the sled is given by:
[tex]F = \mu_{k}N[/tex]
where
N = normal reaction force = mg
where
m = mass of the sled
g = acceleration due to gravity
Thus
[tex]F = \mu_{k}mg[/tex] (3)
Using eqn (1), (2) and (3):
[tex]\mu_{k}mgt = m\Delta v[/tex]
[tex]\mu_{k}gt = v_{A} - v_{B}[/tex]
[tex]t = \frac{v_{A} - v_{B}}{\mu_{k}g}[/tex]
[tex]t = \frac{8.6 - 5.4}{0.25\times 9.8}[/tex]
t = 1.31 s
