let Coefficients of Friction of Rubber on asphalt (dry) =0.7
F= Coefficients of Friction * normal force = 0.7 * 60 =42 N
so the net force of the rubber is zero, meaning it will travel at a constant speed.
When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed
Idk what is growing but if it’s a free than c
Answer:
The SI unit of length verified globally is meters(m)
Answer:
The attached diagram explains the system,
Sum of Fy = 0
N=9.81
N - mgCos60 = 0
F= ukN= (0.53)(9.81) =
F= 5.12 N
So
F.d= 1/2(mv.v) - mgdsin60
-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)
(a) v = 2.436 m/s
For deflection
-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)
by solving for with values of v, m, g, F, k
800x^2 - 11.87 x - 5.938 = 0
by solving the quadratic equation
x = 0.093, -0.079
(b) x = 0.093 m
correct Answer is 0.093m
Explanation: