Answer:
The acceleration of the box is 3 m/s²
Explanation:
Given;
mass of the box, m = 12 kg
horizontal force pulling the box forward, Fx = 48 N
frictional force acting against the box in opposite direction, Fk = 12 N
The net horizontal force on the box, F = 48 N - 12 N
The net horizontal force on the box, F = 36 N
Apply Newton's second law of motion to determine the acceleration of the box;
F = ma
where;
F is the net horizontal force on the box
a is the acceleration of the box
a = F / m
a = 36 / 12
a = 3 m/s²
Therefore, the acceleration of the box is 3 m/s²
Answer:
2.2386 m/s
Explanation:
I think you would use the momentum equation.
m1v1=m2v2
m1= mass of the bag (8.5kg)
v1= speed of bag (6.9 m/s)
m2= mass of bag and cart (26.2 kg)
v2= speed of cart and bag
Plug these numbers all into the equation and you solve for v2 which ends up at 2.2386 m/s
Answer: 5.96m/s
Explanation:
Given the following :
Mass of car (m) = 1500kg
Velocity (V) = 5.25m/s
Forward force of engine = 1250N
Diatance moved = 4.8m
Final Velocity =?
Final kinetic energy = Initial kinetic energy + work done by engine
Initial kinetic energy = 0.5 × mass × velocity^2
Initial kinetic energy = 0.5 × 1500 × 5.25^2
Initial kinetic energy = 20671.875 J
Work done by engine = Force × distance
Work done by engine = 1250 × 4.8 = 6000J
Final kinetic energy = (20671.875 + 6000) J
= 26671.875 J
From kinetic energy = 0.5mv^2
26671.875 = 1/2 × 1500 × v^2
53343.75 = 1500v^2
v^2 = 35.5625
v = sqrt(35.5625)
v = 5.96m/s
im taking test rn nd this question was on there, im saying unbalanced force, if it not correct i will put right answer, but im pretty sure the answer is unblanced