A pickup truck is carrying a 10.0 kgkg toolbox, but the tailgate of the truck is missing, so the box can slide out if it starts
moving. The coefficients of static and kinetic friction between the toolbox and the truck bed are 0.30 and 0.13, respectively. Assume that the truck bed is horizontal. What is the shortest time the truck could take to accelerate from rest to 10.2 m/sm/s without causing the toolbox to slide? A pickup truck is carrying a 10.0 toolbox, but the tailgate of the truck is missing, so the box can slide out if it starts moving. The coefficients of static and kinetic friction between the toolbox and the truck bed are 0.30 and 0.13, respectively. Assume that the truck bed is horizontal. What is the shortest time the truck could take to accelerate from rest to 10.2 without causing the toolbox to slide? 3.47 ss 8.00 ss 2.60 ss 2.25 ss
1 answer:
Answer:
Shortest time = 3.47 seconds
Explanation:
We know that Force of gravity of an object is given by;
F_g = mg
Also, Force due to kinetic friction is given as;
F_friction = μ_k•F_n
Force due to static friction is given as;
F_static = μ_s•F_n
Where ;
μ_k = coefficient of kinetic friction
μ_s = coefficient of static friction
F_n = Normal Force
Let's solve for the static crate.
From the free body diagram i attached, the resultant force in the vertical y direction will be;
N - mg = 0
Thus, N = mg
WherN is the
From earlier, we know that;
F_static = μ_s•F_n
Thus,
F_static = μ_s•mg
Now, in the horizontal x direction,
F_static = ma
Thus,
μ_s•mg = ma
m will cancel out, and;
a = μ_s•g
a = 0.3 x 9.81 = 2.943 m/s²
Not, to get the final the final speed of the tool box, we'll use;
V_f = V_i + a•Δt
Where;
Δt is shortest time.
V_f = final velocity
V_i = initial velocity
Thus, making Δt the subject, we have;
Δt = (V_f - V_i)/a = (10.2 - 0)/2.943 = 3.47 s
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Answer:
(a) Ff = 0.128 N
(b μk = 0.1135
Explanation:
kinematic analysis
Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:
vf=v₀+a*t Formula (1)
d= v₀t+ (1/2)*a*t² Formula (2)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s
Calculation of the acceleration of the hockey puck
We apply the Formula (1)
vf=v₀+a*t v₀=5.8 m/s , vf=0
0=5.8+a*t
-5.8 = a*t
a= -5.8/t Equation (1)
We replace a= -5.8/t in the Formula (2)
d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s
15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²
15.1= 5.8*t-2.9*t
15.1= 2.9*t
t = 15.1 / 2.9
t= 5.2 s
We replace t= 5.2 s in the equation (1)
a= -5.8/5.2
a= -1.115 m/s²
(a) Calculation of the frictional force (Ff)
We apply Newton's second law
∑F = m*a Formula (3)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Look at the free body diagram of the hockey puck in the attached graphic
∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²
-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation
Ff = 0.128 N
(b) Calculation of the coefficient of friction (μk)
N: Normal Force (N)
W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight
g: acceleration due to gravity =9.8 m/s²
∑Fy = 0
N-W=0
N = W
N = 1.127 N
μk = Ff/N
μk = 0.128/1.127
μk = 0.1135
Answer:
(a)
(b)
(c)
(d)
Solution:
As per the question:
Refractive index of medium 1,
Angle of refraction for medium 1,
Angle of refraction for medium 2,
Now,
(a) The expression for the refractive index of medium 2 is given by using Snell's law:
where
= Refractive Index of medium 2
Now,
(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:
(c) To calculate the velocity of light in medium 1:
We know that:
Thus for medium 1
(d) To calculate the velocity of light in medium 2:
For medium 2: