Answer:
3.90 degrees
Explanation:
Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is
W = mg = 30*9.81 = 294.3 N
This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.
We can calculate the parallel one since it's the one that affects the force required to push up
F = WsinΘ
Since customer would not complain if the force is no more than 20N
F = 20
So the ramp cannot be larger than 3.9 degrees
<span>The answer is, 7.44 kg*m/s</span>
Answer:
(a) T= 38.4 N
(b) m= 26.67 kg
Explanation:
We apply Newton's second law:
∑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)
Kinematics
d= v₀t+ (1/2)*a*t² (Formula 2)
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/s²
v₀=0, d=18 m , t=5 s
We apply the formula 2 to calculate the accelerations of the blocks:
d= v₀t+ (1/2)*a*t²
18= 0+ (1/2)*a*(5)²
a= (2*18) / ( 25) = 1.44 m/s²
to the right
We apply Newton's second law to the block A
∑Fx = m*ax
60-T = 15*1.44
60 - 15*1.44 = T
T = 38.4 N
We apply Newton's second law to the block B
∑Fx = m*ax
T = m*ax
38.4 = m*1.44
m= (38.4) / (1.44)
m = 26.67 kg
Nothing at all it can't get
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