Answer:
(A) 10052.2 m/s²
(B) 0.00678 seconds
Explanation:
From the question,
(A) Applying
V² = U²+2as..................... Equation 1
Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.
make a the subject of the equation
a = (V²-U²)/2s........................ Equation 2
Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m
Substitute these values into equation 2
a = (68²-0²)/(2×0.230)
a = 10052.2 m/s²
(B) Using,
a = (V-U)/t......................... Equation 3
Where t= time.
make t the subject of the equation
t = (V-U)/a......................... Equation 4
Given: V = 68 m/s, U = 0 m/s, a = 10052.2
Substitute into equation 4
t = (68-0)/10052.2
t = 0.00678 seconds
Answer:
<em>2.78m/s²</em>
Explanation:
Complete question:
<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>
According to Newton's second law of motion:

Where:
is the coefficient of friction
g is the acceleration due to gravity
Fm is the moving force acting on the body
Ff is the frictional force
m is the mass of the box
a is the acceleration'
Given

Required
acceleration of the box
Substitute the given parameters into the resulting expression above:
Recall that:

9.8sin30 - 0.25(9.8)cos30 = ax
9.8(0.5) - 0.25(9.8)(0.866) = ax
4.9 - 2.1217 = ax
ax = 2.78m/s²
<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>